central net force particle model circular motion problem solving

• 6.2 Uniform Circular Motion
• Introduction
• 1.1 Physics: Definitions and Applications
• 1.2 The Scientific Methods
• 1.3 The Language of Physics: Physical Quantities and Units
• Section Summary
• Key Equations
• Concept Items
• Critical Thinking Items
• Performance Task
• Multiple Choice
• Short Answer
• Extended Response
• 2.1 Relative Motion, Distance, and Displacement
• 2.2 Speed and Velocity
• 2.3 Position vs. Time Graphs
• 2.4 Velocity vs. Time Graphs
• 3.1 Acceleration
• 3.2 Representing Acceleration with Equations and Graphs
• 4.2 Newton's First Law of Motion: Inertia
• 4.3 Newton's Second Law of Motion
• 4.4 Newton's Third Law of Motion
• 5.1 Vector Addition and Subtraction: Graphical Methods
• 5.2 Vector Addition and Subtraction: Analytical Methods
• 5.3 Projectile Motion
• 5.4 Inclined Planes
• 5.5 Simple Harmonic Motion
• 6.1 Angle of Rotation and Angular Velocity
• 6.3 Rotational Motion
• 7.1 Kepler's Laws of Planetary Motion
• 7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity
• 8.1 Linear Momentum, Force, and Impulse
• 8.2 Conservation of Momentum
• 8.3 Elastic and Inelastic Collisions
• 9.1 Work, Power, and the Work–Energy Theorem
• 9.2 Mechanical Energy and Conservation of Energy
• 9.3 Simple Machines
• 10.1 Postulates of Special Relativity
• 10.2 Consequences of Special Relativity
• 11.1 Temperature and Thermal Energy
• 11.2 Heat, Specific Heat, and Heat Transfer
• 11.3 Phase Change and Latent Heat
• 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium
• 12.2 First law of Thermodynamics: Thermal Energy and Work
• 12.3 Second Law of Thermodynamics: Entropy
• 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators
• 13.1 Types of Waves
• 13.2 Wave Properties: Speed, Amplitude, Frequency, and Period
• 13.3 Wave Interaction: Superposition and Interference
• 14.1 Speed of Sound, Frequency, and Wavelength
• 14.2 Sound Intensity and Sound Level
• 14.3 Doppler Effect and Sonic Booms
• 14.4 Sound Interference and Resonance
• 15.1 The Electromagnetic Spectrum
• 15.2 The Behavior of Electromagnetic Radiation
• 16.1 Reflection
• 16.2 Refraction
• 16.3 Lenses
• 17.1 Understanding Diffraction and Interference
• 17.2 Applications of Diffraction, Interference, and Coherence
• 18.1 Electrical Charges, Conservation of Charge, and Transfer of Charge
• 18.2 Coulomb's law
• 18.3 Electric Field
• 18.4 Electric Potential
• 18.5 Capacitors and Dielectrics
• 19.1 Ohm's law
• 19.2 Series Circuits
• 19.3 Parallel Circuits
• 19.4 Electric Power
• 20.1 Magnetic Fields, Field Lines, and Force
• 20.2 Motors, Generators, and Transformers
• 20.3 Electromagnetic Induction
• 21.1 Planck and Quantum Nature of Light
• 21.2 Einstein and the Photoelectric Effect
• 21.3 The Dual Nature of Light
• 22.1 The Structure of the Atom
• 22.2 Nuclear Forces and Radioactivity
• 22.3 Half Life and Radiometric Dating
• 22.4 Nuclear Fission and Fusion
• 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation
• 23.1 The Four Fundamental Forces
• 23.2 Quarks
• 23.3 The Unification of Forces
• A | Reference Tables

Section Learning Objectives

By the end of this section, you will be able to do the following:

• Describe centripetal acceleration and relate it to linear acceleration
• Describe centripetal force and relate it to linear force
• Solve problems involving centripetal acceleration and centripetal force

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.
• (D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Circular and Rotational Motion, as well as the following standards:

Section Key Terms

Centripetal acceleration.

[BL] [OL] Review uniform circular motion. Ask students to give examples of circular motion. Review linear acceleration.

In the previous section, we defined circular motion . The simplest case of circular motion is uniform circular motion , where an object travels a circular path at a constant speed . Note that, unlike speed, the linear velocity of an object in circular motion is constantly changing because it is always changing direction. We know from kinematics that acceleration is a change in velocity , either in magnitude or in direction or both. Therefore, an object undergoing uniform circular motion is always accelerating, even though the magnitude of its velocity is constant.

You experience this acceleration yourself every time you ride in a car while it turns a corner. If you hold the steering wheel steady during the turn and move at a constant speed, you are executing uniform circular motion. What you notice is a feeling of sliding (or being flung, depending on the speed) away from the center of the turn. This isn’t an actual force that is acting on you—it only happens because your body wants to continue moving in a straight line (as per Newton’s first law) whereas the car is turning off this straight-line path. Inside the car it appears as if you are forced away from the center of the turn. This fictitious force is known as the centrifugal force . The sharper the curve and the greater your speed, the more noticeable this effect becomes.

[BL] [OL] [AL] Demonstrate circular motion by tying a weight to a string and twirling it around. Ask students what would happen if you suddenly cut the string? In which direction would the object travel? Why? What does this say about the direction of acceleration? Ask students to give examples of when they have come across centripetal acceleration.

Figure 6.7 shows an object moving in a circular path at constant speed. The direction of the instantaneous tangential velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity; in this case it points roughly toward the center of rotation. (The center of rotation is at the center of the circular path). If we imagine Δ s Δ s becoming smaller and smaller, then the acceleration would point exactly toward the center of rotation, but this case is hard to draw. We call the acceleration of an object moving in uniform circular motion the centripetal acceleration a c because centripetal means center seeking .

Consider Figure 6.7 . The figure shows an object moving in a circular path at constant speed and the direction of the instantaneous velocity of two points along the path. Acceleration is in the direction of the change in velocity and points toward the center of rotation. This is strictly true only as Δ s Δ s tends to zero.

Now that we know that the direction of centripetal acceleration is toward the center of rotation, let’s discuss the magnitude of centripetal acceleration. For an object traveling at speed v in a circular path with radius r , the magnitude of centripetal acceleration is

Centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you may have noticed when driving a car, because the car actually pushes you toward the center of the turn. But it is a bit surprising that a c is proportional to the speed squared. This means, for example, that the acceleration is four times greater when you take a curve at 100 km/h than at 50 km/h.

We can also express a c in terms of the magnitude of angular velocity . Substituting v = r ω v = r ω into the equation above, we get a c = ( r ω ) 2 r = r ω 2 a c = ( r ω ) 2 r = r ω 2 . Therefore, the magnitude of centripetal acceleration in terms of the magnitude of angular velocity is

Tips For Success

The equation expressed in the form a c = rω 2 is useful for solving problems where you know the angular velocity rather than the tangential velocity.

Virtual Physics

Ladybug motion in 2d.

In this simulation, you experiment with the position, velocity, and acceleration of a ladybug in circular and elliptical motion. Switch the type of motion from linear to circular and observe the velocity and acceleration vectors. Next, try elliptical motion and notice how the velocity and acceleration vectors differ from those in circular motion.

Grasp Check

In uniform circular motion, what is the angle between the acceleration and the velocity? What type of acceleration does a body experience in the uniform circular motion?

• The angle between acceleration and velocity is 0°, and the body experiences linear acceleration.
• The angle between acceleration and velocity is 0°, and the body experiences centripetal acceleration.
• The angle between acceleration and velocity is 90°, and the body experiences linear acceleration.
• The angle between acceleration and velocity is 90°, and the body experiences centripetal acceleration.

Centripetal Force

[BL] [OL] [AL] Using the same demonstration as before, ask students to predict the relationships between the quantities of angular velocity, centripetal acceleration, mass, centripetal force. Invite students to experiment by using various lengths of string and different weights.

Because an object in uniform circular motion undergoes acceleration (by changing the direction of motion but not the speed), we know from Newton’s second law of motion that there must be a net external force acting on the object. Since the magnitude of the acceleration is constant, so is the magnitude of the net force, and since the acceleration points toward the center of the rotation, so does the net force.

Any force or combination of forces can cause a centripetal acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, the friction between a road and the tires of a car as it goes around a curve, or the normal force of a roller coaster track on the cart during a loop-the-loop.

The component of any net force that causes circular motion is called a centripetal force . When the net force is equal to the centripetal force, and its magnitude is constant, uniform circular motion results. The direction of a centripetal force is toward the center of rotation, the same as for centripetal acceleration. According to Newton’s second law of motion, a net force causes the acceleration of mass according to F net = m a . For uniform circular motion, the acceleration is centripetal acceleration: a = a c . Therefore, the magnitude of centripetal force, F c , is F c = m a c F c = m a c .

By using the two different forms of the equation for the magnitude of centripetal acceleration, a c = v 2 / r a c = v 2 / r and a c = r ω 2 a c = r ω 2 , we get two expressions involving the magnitude of the centripetal force F c F c . The first expression is in terms of tangential speed, the second is in terms of angular speed: F c = m v 2 r F c = m v 2 r and F c = m r ω 2 F c = m r ω 2 .

Both forms of the equation depend on mass, velocity, and the radius of the circular path. You may use whichever expression for centripetal force is more convenient. Newton’s second law also states that the object will accelerate in the same direction as the net force. By definition, the centripetal force is directed towards the center of rotation, so the object will also accelerate towards the center. A straight line drawn from the circular path to the center of the circle will always be perpendicular to the tangential velocity. Note that, if you solve the first expression for r , you get

From this expression, we see that, for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Watch Physics

Centripetal force and acceleration intuition.

This video explains why centripetal force, when it is equal to the net force and has constant magnitude, creates centripetal acceleration and uniform circular motion.

Misconception Alert

Some students might be confused between centripetal force and centrifugal force. Centrifugal force is not a real force but the result of an accelerating reference frame, such as a turning car or the spinning Earth. Centrifugal force refers to a fictional center fleeing force.

• The yoyo will fly inward in the direction of the centripetal force.
• The yoyo will fly outward in the direction of the centripetal force.
• The yoyo will fly to the left in the direction of the tangential velocity.
• The yoyo will fly to the right in the direction of the tangential velocity.

Solving Centripetal Acceleration and Centripetal Force Problems

To get a feel for the typical magnitudes of centripetal acceleration, we’ll do a lab estimating the centripetal acceleration of a tennis racket and then, in our first Worked Example, compare the centripetal acceleration of a car rounding a curve to gravitational acceleration. For the second Worked Example, we’ll calculate the force required to make a car round a curve.

Estimating Centripetal Acceleration

In this activity, you will measure the swing of a golf club or tennis racket to estimate the centripetal acceleration of the end of the club or racket. You may choose to do this in slow motion. Recall that the equation for centripetal acceleration is a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 .

• One tennis racket or golf club
• One ruler or tape measure
• Work with a partner. Stand a safe distance away from your partner as he or she swings the golf club or tennis racket.
• Describe the motion of the swing—is this uniform circular motion? Why or why not?
• Try to get the swing as close to uniform circular motion as possible. What adjustments did your partner need to make?
• Measure the radius of curvature. What did you physically measure?
• By using the timer, find either the linear or angular velocity, depending on which equation you decide to use.
• What is the approximate centripetal acceleration based on these measurements? How accurate do you think they are? Why? How might you and your partner make these measurements more accurate?

The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error. The swing of the golf club or racket can be made very close to uniform circular motion. For this, the person would have to move it at a constant speed, without bending their arm. The length of the arm plus the length of the club or racket is the radius of curvature. Accuracy of measurements of angular velocity and angular acceleration will depend on resolution of the timer used and human observational error.

Was it more useful to use the equation a c = v 2 r a c = v 2 r or a c = r ω 2 a c = r ω 2 in this activity? Why?

• It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be easier.
• It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be easier.
• It should be simpler to use a c = r ω 2 a c = r ω 2 because measuring angular velocity through observation would be difficult.
• It should be simpler to use a c = v 2 r a c = v 2 r because measuring tangential velocity through observation would be difficult.

Worked Example

Comparing centripetal acceleration of a car rounding a curve with acceleration due to gravity.

A car follows a curve of radius 500 m at a speed of 25.0 m/s (about 90 km/h). What is the magnitude of the car’s centripetal acceleration? Compare the centripetal acceleration for this fairly gentle curve taken at highway speed with acceleration due to gravity ( g ).

Because linear rather than angular speed is given, it is most convenient to use the expression a c = v 2 r a c = v 2 r to find the magnitude of the centripetal acceleration.

Entering the given values of v = 25.0 m/s and r = 500 m into the expression for a c gives

To compare this with the acceleration due to gravity ( g = 9.80 m/s 2 ), we take the ratio a c / g = ( 1.25  m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 a c / g = ( 1.25  m/s 2 ) / ( 9.80 m/s 2 ) = 0.128 . Therefore, a c = 0.128 g a c = 0.128 g , which means that the centripetal acceleration is about one tenth the acceleration due to gravity.

Frictional Force on Car Tires Rounding a Curve

• Calculate the centripetal force exerted on a 900 kg car that rounds a 600-m-radius curve on horizontal ground at 25.0 m/s.
• Static friction prevents the car from slipping. Find the magnitude of the frictional force between the tires and the road that allows the car to round the curve without sliding off in a straight line.
• If the car would slip if it were to be traveling any faster, what is the coefficient of static friction between the tires and the road? Could we conclude anything about the coefficient of static friction if we did not know whether the car could round the curve any faster without slipping?

Strategy and Solution for (a)

We know that F c = m v 2 r F c = m v 2 r . Therefore,

Strategy and Solution for (b)

The image above shows the forces acting on the car while rounding the curve. In this diagram, the car is traveling into the page as shown and is turning to the left. Friction acts toward the left, accelerating the car toward the center of the curve. Because friction is the only horizontal force acting on the car, it provides all of the centripetal force in this case. Therefore, the force of friction is the centripetal force in this situation and points toward the center of the curve.

Strategy and Solution for (c)

If the car is about to slip, the static friction is at its maximum value and f = μ s N = μ s m g f = μ s N = μ s m g . Solving for μ s μ s , we get μ s = 938 900 × 9.8 = 0.11 μ s = 938 900 × 9.8 = 0.11 . Regardless of whether we know the maximum allowable speed for rounding the curve, we can conclude this is a minimum value for the coefficient.

Since we found the force of friction in part (b), we could also solve for the coefficient of friction, since f = μ s N = μ s m g f = μ s N = μ s m g . The static friction is only equal to μ s N μ s N when it is at the maximum possible value. If the car could go faster, the friction at the given speed would still be the same as we calculated, but the coefficient of static friction would be larger.

Practice Problems

Calculate the centripetal acceleration of an object following a path with a radius of a curvature of 0.2 m and at an angular velocity of 5 rad/s.

Check Your Understanding

• Uniform circular motion is when an object accelerates on a circular path at a constantly increasing velocity.
• Uniform circular motion is when an object travels on a circular path at a variable acceleration.
• Uniform circular motion is when an object travels on a circular path at a constant speed.
• Uniform circular motion is when an object travels on a circular path at a variable speed.

Which of the following is centripetal acceleration?

• The acceleration of an object moving in a circular path and directed radially toward the center of the circular orbit
• The acceleration of an object moving in a circular path and directed tangentially along the circular path
• The acceleration of an object moving in a linear path and directed in the direction of motion of the object
• The acceleration of an object moving in a linear path and directed in the direction opposite to the motion of the object
• Yes, the object is accelerating, so a net force must be acting on it.
• Yes, because there is no acceleration.
• No, because there is acceleration.
• No, because there is no acceleration.

Identify two examples of forces that can cause centripetal acceleration.

• The force of Earth’s gravity on the moon and the normal force
• The force of Earth’s gravity on the moon and the tension in the rope on an orbiting tetherball
• The normal force and the force of friction acting on a moving car
• The normal force and the tension in the rope on a tetherball

Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the formative assessment will help identify which objective is causing the problem and direct students to the relevant content.

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6.3: Uniform circular motion

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As we saw in Chapter 4, “uniform circular motion” is defined to be motion along a circle with constant speed. This may be a good time to review Section 4.4 for the kinematics of motion along a circle. In particular, for the uniform circular motion of an object around a circle of radius \(R\) , you should recall that:

• The velocity vector, \(\vec v\) , is always tangent to the circle.
• The acceleration vector, \(\vec a\) , is always perpendicular to the velocity vector, because the magnitude of the velocity vector does not change.
• The acceleration vector, \(\vec a\) , always points towards the center of the circle.
• The acceleration vector has magnitude \(a=v^2/R\) .
• The angular velocity, \(\omega\) , is related to the magnitude of the velocity vector by \(v=\omega R\) and is constant.
• The angular acceleration, \(\alpha\) , is zero for uniform circular motion, since the angular velocity does not change.

In particular, you should recall that even if the speed is constant, the acceleration vector is always non-zero in uniform circular motion because the velocity changes direction . According to Newton’s Second Law, this implies that there must be a net force on the object that is directed towards the center of the circle 1 (parallel to the acceleration): \[\begin{aligned} \sum \vec F = m\vec a\end{aligned}\] where the acceleration has a magnitude \(a=v^2/R\) . Because the acceleration is directed towards the center of the circle, we sometimes call it a “radial” acceleration (parallel to the radius), \(a_R\) , or a “centripetal” acceleration (directed towards the center), \(a_c\) .

Consider an object in uniform circular motion in a horizontal plane on a frictionless surface, as depicted in Figure \(\PageIndex{1}\).

The only way for the object to undergo uniform circular motion as depicted is if the net force on the object is directed towards the center of the circle. One way to have a force that is directed towards the center of the circle is to attach a string between the center of the circle and the object, as shown in Figure \(\PageIndex{1}\). If the string is under tension, the force of tension will always be towards the center of the circle. The forces on the object are thus:

• \(\vec F_g\) , its weight with magnitude \(mg\) .
• \(\vec N\) , a normal forced exerted by the surface.
• \(\vec T\) , a force of tension exerted by the string.

The forces are depicted in the free-body diagram shown in Figure \(\PageIndex{2}\) (as viewed from the side), where we also drew the acceleration vector. Note that this free-body diagram is only “valid” at a particular instant in time since the acceleration vector continuously changes direction and would not always be lined up with the \(x\) axis.

Writing out the \(x\) and \(y\) components of Newton’s Second Law: \[\begin{aligned} \sum F_x &= T = ma_R\\ \sum F_y &= N - F_g =0\end{aligned}\] The \(y\) component just tells us that the normal force must have the same magnitude as the weight because the object is not accelerating in the vertical direction. The \(x\) component tells us the relation between the magnitudes of the tension in the string and the radial acceleration. Using the speed of the object, we can also write the relation between the tension and the speed: \[\begin{aligned} T &= ma_R=m\frac{v^2}{R}\\\end{aligned}\] Thus, we find that the tension in the string increases with the square of the speed, and decreases with the radius of the circle.

Exercise \(\PageIndex{1}\)

An object is undergoing uniform circular motion in the horizontal plane, when the string connecting the object to the center of rotation suddenly breaks. What path will the block take after the string broke?

Example \(\PageIndex{1}\)

A car goes around a curve which can be approximated as the arc of a circle of radius \(R\), as shown in Figure \(\PageIndex{4}\). The coefficient of static friction between the tires of the car and the road is \(µ_{s}\). What is the maximum speed with which the car can go around the curve without skidding?

If the car is going at constant speed around a circle, then the sum of the forces on the car must be directed towards the center of the circle. The only force on the car that could be directed towards the center of the circle is the force of friction between the tires and the road. If the road were perfectly slick (think driving in icy conditions), it would not be possible to drive around a curve since there could be no force of friction. The forces on the car are:

• \(\vec N\) , a normal force exerted upwards by the road.
• \(\vec f_s\) , a force of static friction between the tires and the road. This is static friction, because the surface of the tire does not move relative to the surface of the road if the car is not skidding. The force of static friction has a magnitude that is at most \(f_s\leq\mu_sN\) .

The forces on the car are shown in the free-body diagram in Figure 6.3.5.

The \(y\) component of Newton’s Second Law tells us that the normal force exerted by the road must equal the weight of the car: \[\begin{aligned} \sum F_y = N-F_g&=0\\ \therefore N &=mg\end{aligned}\] The \(x\) component relates the force of friction to the radial acceleration (and thus to the speed): \[\begin{aligned} \sum F_x = f_s =ma_R&=m\frac{v^2}{R}\\ \therefore f_s &= m\frac{v^2}{R}\end{aligned}\] The force of friction must be less than or equal to \(f_s\leq\mu_sN=\mu_smg\) (since \(N=mg\) from the \(y\) component of Newton’s Second Law), which gives us a condition on the speed: \[\begin{aligned} f_s = m\frac{v^2}{R}&\leq\mu_smg\\ v^2 &\leq \mu_s g R\\ \therefore v &\leq \sqrt{\mu_s g R}\end{aligned}\] Thus, if the speed is less than \(\sqrt{\mu_s g R}\) , the car will not skid and the magnitude of the force of static friction, which results in an acceleration towards the center of the circle, will be smaller or equal to its maximal possible value.

Discussion:

The model for the maximum speed that the car can travel around the curve makes sense because:

• The dimension of \(\sqrt{\mu_s g R}\) is speed.
• The speed is larger if the radius of the curve is larger (one can go faster around a wider curve without skidding).
• The speed is larger if the coefficient of friction is large (if the force of friction is larger, a larger radial acceleration can be sustained).

Example \(\PageIndex{2}\)

A ball is attached to a mass-less string and executing circular motion along a circle of radius \(R\) that is in the vertical plane, as depicted in Figure \(\PageIndex{6}\). Can the speed of the ball be constant? What is the minimum speed of the ball at the top of the circle if it is able to make it around the circle?

The forces that are acting on the ball are:

Figure \(\PageIndex{7}\) shows the free-body diagram for the forces on the ball at three different locations along the path of the circle.

In order for the ball to go around in a circle, there must be at least a component of the net force on the ball that is directed towards the center of the circle at all times. In the bottom half of the circle (positions 1 and 2), only the tension can have a component directed towards the center of the circle.

Consider in particular the position labeled 2, when the string is horizontal and the tension is equal to \(\vec T_2\) . The free-body diagram in Figure \(\PageIndex{7}\) also shows the vector sum of the weight and tension at position 2 (the red arrow labeled \(\sum \vec F\) ), which points downwards and to the left. It is thus clearly impossible for the acceleration vector to point towards the center of the circle, and the acceleration will have components that are both tangential ( \(a_T\) ) to the circle and radial ( \(a_R\) ), as shown by the vector \(\vec a_2\) in Figure \(\PageIndex{7}\).

The radial component of the acceleration will change the direction of the velocity vector so that the ball remains on the circle, and the tangential component will reduce the magnitude of the velocity vector. According to our model, it is thus impossible for the ball to go around the circle at constant speed, and the speed must decrease as it goes from position 2 to position 3, no matter how one pulls on the string (you can convince yourself of this by drawing the free-body diagram at any point between points 2 and 3).

The minimum speed for the ball at the top of the circle is given by the condition that the tension in the string is zero just at the top of the trajectory (position 3). The ball can still go around the circle because, at position 3, gravity is towards the center of the circle and can thus give an acceleration that is radial, even with no tension. The \(y\) component of Newton’s Second Law, at position 3 gives: \[\begin{aligned} \sum F_y = -F_g &= ma_y\\ \therefore a_y &=-g\end{aligned}\] The magnitude of the acceleration is the radial acceleration, and is thus related to the speed at the top of the trajectory: \[\begin{aligned} a_R&=-a_y=g = m\frac{v^2}{R}\\ \therefore v_{min}&=\sqrt{\frac{gR}{m}}\end{aligned}\] which is the minimum speed at the top of the trajectory for the ball to be able to continue along the circle. The tension in the string would change as the ball moves around the circle, and will be highest at the bottom of the trajectory, since the tension has to be bigger than gravity so that the net force at the bottom of the trajectory is upwards (towards the center of the circle).

The model for the minimum speed of the ball at the top of the circle makes sense because:

• \(\sqrt{\frac{gR}{m}}\) has the dimension of speed.
• The minimum velocity is larger if the circle has a larger radius (try this with a mass attached at the end of a string).
• The minimum velocity is larger if the mass is bigger (again, try this at home!).

Exercise \(\PageIndex{2}\)

Consider a ball attached to a string, being spun in a vertical circle (such as the one depicted in Figure \(\PageIndex{6}\)). If you shortened the string, how would the minimum angular velocity (measured at the top of the trajectory) required for the ball to make it around the circle change?

• It would decrease
• It would stay the same
• It would increase

Banked curves

As we saw in Example 6.3.1 , there is a maximum speed with which a car can go around a curve before it starts to skid. You may have noticed that roads, highways especially, are banked where there are curves. Racetracks for cars that go around an oval (the boring kind of car races) also have banked curves. As we will see, this allows the speed of vehicles to be higher when going around the curve; or rather, it makes the curves safer as the speed at which vehicles would skid is higher. In Example 6.3.1 , we saw that it was the force of static friction between the tires of the car and the road that provided the only force with a component towards the center of the circle. The idea of using a banked curve is to change the direction of the normal force between the road and the car tires so that it, too, has a component in the direction towards the center of the circle.

Consider the car depicted in Figure \(\PageIndex{8}\) which is seen from behind making a left turn around a curve that is banked by an angle \(\theta\) with respect to the horizontal and can be modeled as an arc from a circle of radius \(R\) .

The forces exerted on the car are the same as in Example 6.3.1 , except that they point in different directions. The forces are:

• \(\vec N\) , a normal force exerted by the road, perpendicular to the surface of the road.
• \(\vec f_s\) , a force of static friction between the tires and the road. This is static friction, because the surface of the tire does not move relative to the surface of the road if the car is not skidding. The force of static friction has a magnitude that is at most \(f_s\leq\mu_sN\) and is perpendicular to the normal force. The force could be either upwards or downwards, depending on the other forces on the car .

A free-body diagram for the forces on the car is shown in Figure 6.3.9, along with the acceleration (which is in the radial direction, towards the center of the circle), and our choice of coordinate system (choosing \(x\) parallel to the acceleration). The direction of the force of static friction is not known a priori and depends on the speed of the car:

• If the speed of the car is zero, the force of static friction is upwards. With a speed of zero, the radial acceleration is zero, and the sum of the forces must thus be zero. The impeding motion of the car would be to slide down the banked curve (just like a block on an incline).
• If the speed of the car is very large, the force of static friction is downwards, as the impeding motion of the car would be to slide up the bank. The natural motion of the car is to go in a straight line (Newton’s First Law). If the components of the normal force and of the force of static friction directed towards the center of the circle are too small to allow the car to turn, then the car would slide up the bank (so the impeding motion is up the bank and the force of static friction is downwards).

There is thus an “ideal speed” at which the force of static friction is precisely zero, and the \(x\) component of the normal force is responsible for the radial acceleration. At higher speeds, the force of static friction is downwards and increases in magnitude to keep the car’s acceleration towards the center of the circle. At some maximal speed, the force of friction will reach its maximal value, and no longer be able to keep the car’s acceleration pointing towards the center of the circle. At speeds lower than the ideal speed, the force of friction is directed upwards to prevent the car from sliding down the bank. If the coefficient of static friction is too low, it is possible that at low speeds, the car would start to slide down the bank (so there would be a minimum speed below which the car would start to slide down).

Let us model the situation where the force of static friction is identically zero so that we can determine the ideal speed for the banked curve. The only two forces on the car are thus its weight and the normal force. The \(x\) and \(y\) component of Newton’s Second Law give:

\[\begin{aligned} \label{eq:applyingnewtonslaws:carbank_x} \sum F_x &= N\sin\theta = ma_R=m\frac{v^2}{R}\nonumber\end{aligned}\]

\[\therefore N\sin\theta = m\frac{v^{2}}{R}\]

\[\begin{aligned} \label{eq:applyingnewtonslaws:carbank_y} \sum F_y &= N\cos\theta-F_g = 0\nonumber\end{aligned}\]

\[\therefore N\cos\theta =mg\]

We can divide Equation 6.3.1 by Equation 6.3.2 , noting that \(\tan\theta=\sin\theta/\cos\theta\) , to obtain: \[\begin{aligned} \tan\theta &= \frac{v^2}{gR}\\ \therefore v_{ideal} &=\sqrt{gR\tan\theta}\end{aligned}\] At this speed, the force of static friction is zero. In practice, one would use this equation to determine which bank angle to use when designing a road, so that the ideal speed is around the speed limit or the average speed of traffic. We leave it as an exercise to determine the maximal speed that the car can go around the curve before sliding out.

Inertial forces in circular motion

As you sit in a car that is going around a curve, you will feel pushed outwards, away from the center of the circle that the car is going around. This is because of your inertia (Newton’s First Law), and your body would go in a straight line if the car were not exerting a net force on you towards the center of the circle. You are not so much feeling a force that is pushing you outwards as you are feeling the effects of the car seat pushing you inwards; if you were leaning against the side of the car that is on the outside of the curve, you would feel the side of the car pushing you inwards towards the center of the curve, even if it “feels” like you are pushing outwards against the side of the car.

If we model your motion looking at you from the ground, we would include a force of friction between the car seat (or the side of the car, or both) and you that is pointing towards the center of the circle, so that the sum of the forces exerted on you is towards the center of the circle. We can also model your motion from the non-inertial frame of the car. As you recall, because this is a non-inertial frame of reference, we need to include an additional inertial force, \(\vec F_I\) , that points opposite of the acceleration of the car, with magnitude \(F_I=ma_R\) (if the net acceleration of the car is \(a_R\) ). Inside the non-inertial frame of reference of the car, your acceleration (relative to the reference frame, i.e. the car) is zero. This is illustrated by the diagrams in Figure \(\PageIndex{10}\).

The \(y\) component of Newton’s Second Law in both frames of reference is the same: \[\begin{aligned} \sum F_y&=N-F_g=0\\ \therefore N&=mg\end{aligned}\] and simply tells us that the normal force is equal to the weight. In the reference frame of the ground, the \(x\) component of Newton’s Second Law gives: \[\begin{aligned} \sum F_x &= f_s = ma_R\\ \therefore f_s &= m\frac{v^2}{R}\end{aligned}\] In the frame of reference of the car, where your acceleration is zero and an inertial force of magnitude \(F_I=mv^2/R\) is exerted on you, the \(x\) component of Newton’s Second Law gives: \[\begin{aligned} \sum F_x &= f_s-F_I = 0\\ \therefore f_s - m\frac{v^2}{R} &= 0\end{aligned}\] which of course, mathematically, is exactly equivalent. The inertial force is not a real force in the sense that it is not exerted by anything. It only comes into play because we are trying to use Newton’s Laws in a non-inertial frame of reference. However, it does provide a good model for describing the sensation that we have of being pushed outwards when the car goes around a curve. Sometimes, people will refer to this force as a “centrifugal” force, which means “a force that points away from the center”. You should however remember that this is not a real force exerted on the object, but is the result of modeling motion in a non-inertial frame of reference.

Exercise \(\PageIndex{3}\)

Jamie is driving his tricycle around a circular pond. Jamie feels a centrifugal force with magnitude \(F_I\) . If Jamie pedals twice as fast, what will be the magnitude of the centrifugal force that he experiences?

• \(\sqrt{2}F_I\)
• \(\frac{1}{2}F_I\)

1. The sum of the forces is often called the “net force” on an object, and in the specific case of uniform circular motion, that net force is sometimes called the “centripetal force” - however, it is not a force in and of itself and it is always the sum of the forces that points towards the center of the circle.

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Uniform Circular Motion Problems with Answers

In the following, some problems on uniform circular motion are provided with detailed answers. These questions on circular motion are beneficial for students in high school and college.

Circular Motion Problems: Kinematic

Problem (1): A 5-kg object moves around a circular track with a radius of 18 cm at a constant speed of 6 m/s. Find  (a) The magnitude and direction of the acceleration of the object. (b) The net force acting upon the object causing this acceleration.

Solution : When an object moves around a circular path at a constant speed, the only acceleration it experiences is centripetal acceleration or radial acceleration.  

This kind of acceleration is always toward the center of the circle, and its magnitude is found by the following formula: \[a_c=\frac{v^2}{r}\] where $v$ is the constant speed with which the object revolves the circle, and $r$ is the radius of the circle. 

(a) The track is circular and the speed of the object is constant, so a centripetal acceleration directed toward the center is applied to the object, whose magnitude is as follows \[a_c=\frac{6^2}{0.18\,{\rm m}}=50\,{\rm \frac{m}{s^2}}\] In the figure below, a top view of the motion is sketched. 

Note : At each point along the circular path, the instantaneous velocity of the revolving object is tangent to the path. The direction of this velocity changes, but its magnitude remains constant. 

Problem (2): On a merry-go-round moving at a speed of 3 m/s, there is a 25-kg child sitting 3 m from the center. Calculate the following: (a) the centripetal acceleration of the child, (b) the net horizontal force acting on the child,  (c) compare the above force with the child's weight.

Solution : (a) The child has a circular motion with a centripetal acceleration given by $a_c=v^2/r$, where $v$ is the constant speed of the revolving object. Therefore, we have \[a_c=\frac{3^2}{3}=3\quad {\rm \frac{m}{s^2}}\] (b) The net force can be found using Newton's second law as $F_{net}=ma_c$, which gives us \[F_{net}=25\times 3=75\quad {\rm N}\] This force is in the same direction as the acceleration, toward the center of the circle.

(c) Weight is calculated as mass times the gravitational acceleration at that place ($g$) or $w=mg$. The ratio between these two forces is \[\frac{F_{net}}{w}=\frac{75}{25\times 10}=0.3\]

Problem (3): A cyclist is moving around a circular path with a radius of 50 m at a speed of 10 m/s. (a) Find the cyclist's acceleration. (b) Considering the combined mass of the cyclist and bicycle to be 120 kg, what is the net force applied to them?

Solution: (a) When moving with constant speed around a circular path, the cyclist experiences a radially inward acceleration known as centripetal acceleration. The magnitude of this acceleration can be found using the formula: \[a_c=\frac{v^2}{r}=\frac{10^2}{50}=2\quad {\rm \frac{m}{s^2}}\] (b) The net force acting on the cyclist and bicycle to keep them moving along the circular track is called the centripetal force. Its magnitude can be determined using Newton's second law: \[F_{net}=ma_c=120\times 2=240\quad N\]

Problem: A particle moves in a uniform circular motion with a radius of 3 cm. The net force radially acting on the object is 30 N. If the mass of the object is 6 kg,  (a) What is centripetal acceleration? (b) What is the particle's linear speed?  (c) What is the acceleration if the radius is doubled without changing the particle's speed? (d) What is the acceleration if the angular velocity is doubled without changing the circle's path?

Solution : Centripetal acceleration is defined as the acceleration that an object experiences when it turns around a circular path. This acceleration points radially inward toward the center of the circular path and is related to the corresponding centripetal force, $F_c$, by the following formula \[F_c=ma_c=m\frac{v^2}{r}\] where $r$ is the radius of the circular path and $v$ is the linear speed of the revolving object. 

(a) According to the above explanation, $a_c$ can be found as \begin{gather*} F_c=ma_c \\ 30=6\times a_c \\ \Rightarrow a_c=5\,\rm m/s^2\end{gather*}  (b) Centripetal acceleration and linear speed are related by $a_c= \frac{v^2}{r}$. Substituting the numerical values into that and solving for $v$ get \begin{gather*} a_c=\frac{v^2}{r} \\\\ 5=\frac{v^2}{0.03} \\\\ \rightarrow v^2=0.15 \\\\ \Rightarrow \boxed{v=\sqrt{0.15}\,\rm m/s} \end{gather*}  (c) According to the centripetal acceleration formula, $a_c=\frac{v^2}{r}$, acceleration is inversely proportional to the radius of the circular path, $a_c\propto 1/r$. Therefore, we have \begin{gather*} \frac{a_c}{a'_c}=\frac{r'}{r} \\\\ \frac{5}{a'_c}=\frac{2\times 3}{3} \\\\ \Rightarrow \boxed{a'_c=2.5\,\rm m/s^2} \end{gather*}  (d) We can also express the centripetal acceleration $a_c$ in terms of the angular velocity $\omega$ using formula \[a_c=r\omega^2\] Substituting the given numerical values into that equation gives \begin{gather*} a_c=r\omega^2 \\ 5=0.03 \times \omega^2 \\\\ \omega=\sqrt{\frac{5}{0.03}}\ \rm rad/s \end{gather*} 

Problem (4): A jet plane is maneuvering in a circular path with a radius of 5.5 km at a constant speed of 2160 km/h. What is the acceleration of the plane in g's?

Solution : This is a circular motion problem with constant speed, so the only acceleration is due to centripetal acceleration.

Because we are going to compare this acceleration with gravitational acceleration, $g=9.8\,{\rm m/s^2}$, it is better to first convert all units into SI units. 

This conversion for constant speed is as follows \[\rm{1\,\frac{km}{h}}=\rm{\frac{1000\,m}{3600\,s}}=\frac{10}{36} {\rm \frac{m}{s}}\] So, 2160 km/h in SI units is converted as \[2160\times \frac{10}{36}=600\quad {\rm \frac{m}{s}}\] Therefore, the magnitude of the centripetal acceleration is \[a_c=\frac{v^2}{r}=\frac{600^2}{5000\,m}=72\quad {\rm \frac{m}{s^2}}\] To compare it with gravity, we should divide it by $g$ \[\frac{a_c}{g}=\frac{72}{9.8}=7.35\] Thus, this acceleration is about seven times greater than the gravitational acceleration, $a_c=7.35\,g$. 

Problem (5): A race car is traveling around a circular path with a radius of 400 m at a speed of 50 m/s. Find the centripetal acceleration of the car.

Solution : In a circular motion with constant speed, the centripetal acceleration is $a_c=v^2/r$. Therefore, substituting the numerical values into this equation, we have \[a_c=\frac{v^2}{r}=\frac{50^2}{400}=6.25\,{\rm \frac{m}{s^2}}\] This acceleration is toward the center of the circle.

Problem (6): A crankshaft of radius 8 cm rotates at 2400 rpm (revolutions per minute). What is the speed of a point at its surface? 

Solution : The aim of this circular motion problem is the conversion of rpm into SI units of speed (m/s), which is done as below: \begin{align*} 1 \frac{rev}{minute}&=\frac{\text{circle's circumference}}{60\,\text{seconds}}\\\\&=\frac{2\pi r}{60}\\\\&=\frac{\pi r}{30}\end{align*} Hence, to convert rpm to $m/s$, we can use the following conversion ratio \[ \rm{1\,rpm}=\frac{\pi r}{30}\,{\rm \frac ms}\] In this problem, the circular motion has a radius of $r=0.08\,{\rm m}$ and an angular speed of 2400 rpm, so we get \[2400\,{\rm rpm}=2400\times\frac{\pi(0.08)}{30}=20.1\,{\rm \frac ms}\]

Problem (7): A 20-kg child sits in a cart to which a 2-m rope is attached. The rope is tied to a motor that rotates the cart. At the instant that the tension in the rope is 100 N, how many revolutions per minute does the cart make?

Solution : This circular motion question aims to find the method of conversion of $m/s$ into $\rm rpm$. 

The child revolves in a circular path, so the net centripetal force it experiences is determined by $F_c=\frac{mv^2}{r}$.

First, use the above equation and solve for the constant speed $v$ as below \[v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{2\times 100}{20}}=\sqrt{10}\,{\rm \frac ms}\] Next, convert the above linear speed in ($\rm \frac ms$) into angular speed in revolutions per minute ($\rm rpm$). 

The simple relation between the two speeds is as $v=r\omega$, where $\omega$ is the angular speed measured in radians per second. \[\omega=\frac{v}{r}=\frac{\sqrt{10}}{2} \,{\rm \frac {rad}{s}}\] Next, as shown below, convert $\rm \frac{rad}{s}$ into $\rm rpm$ \[  \frac{1\,rad}{s}=\frac{\frac{1}{2\pi}\,rev}{\frac {1}{60}\,min}=\frac {30}{\pi}\, \rm rpm\] In the above, we used this fact that each revolution around a circle is equal to $2\pi$ radians. As a result, we have \[\sqrt{10}\,{\rm \frac ms}=\sqrt{10}\left(\frac{30}{\pi}\right)\,{\rm rpm}=30.2\,{\rm rpm}\] Hence, the cart makes about 30 revolutions per minute.

Problem (8): A radially inward constant force of 300 N is exerted on a 2-kg ball as it revolves around a circle of radius of 85 cm. What is the speed of the ball?

Solution : The ball revolves around a circle, so the only radially inward force acting upon the object is the centripetal force, whose magnitude is $F_c=mv^2/r$. Therefore, putting the numerical values into it and solving for the unknown $v$, we have \[v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{0.85\times 300}{2}}=11.3\,{\rm \frac{m}{s}}\]

Problem (9): A 35-kg child makes a turn that is a portion of a circle with a radius of 12 meters. She covers one-quarter of the circular path in 1.6 seconds. Determine the speed, acceleration, and net force applied to the child. 

Solution : The child turns one-quarter of a circle in $1.6$ seconds. 

Using the definition of speed, as distance traveled divided by time elapsed, we can find its speed. 

But here, the distance traveled is the circumference of one-quarter of a circle of radius $12$ meters, so $L=\frac 14 (2\pi r)=\frac{\pi r}{2}$. \begin{align*} speed&=\frac{distance}{\text{time elapsed}}\\\\&=\frac{\pi r/2}{time}\\\\&=\frac{\pi\times (12/2)}{1.6}\\\\&=11.8\quad {\rm m/s}\end{align*} Because the child is moving around a portion of a circular path, centripetal acceleration is applied to the child, whose magnitude is $a_c=v^2/r$. \[a_c=\frac{v^2}{r}=\frac{(11.8)^2}{12}=11.6\quad{\rm \frac{m}{s^2}}\] The net force is also found using Newton's second law of motion as $F_{net}=ma_c$. \[F_{net}=ma_c=35\times 11.6=406\quad {\rm N}\]

Problem (10): A ball of mass 2 kg is attached to a rope having a breaking strength of 1500 N and is whirled around a horizontal circle with a radius of 85 cm. Calculate the maximum speed that the ball can have.

Solution : There is a circular motion, so the only radially inward force that acted upon the ball is a centripetal force whose magnitude is $F_c=mv^2/r$ where $r$ is the radius of the circular path. 

Putting everything into this equation and solving for the unknown velocity, we have \[v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{0.85\times 1500}{2}}=25.25\,{\rm \frac{m}{s}}\]

The following circular motion questions are helpful for the AP physics exam.

Problem (11): The speed of a 515-kg roller-coaster at the bottom of a loop of radius 10 m is 20 m/s. Find the net vertical force pushing up on the object at this point of the circular path.

Solution: Note that because of gravity, the speed of the object changes at each point of the track, so the object is not undergoing uniform circular motion. Nonetheless, we can use the centripetal acceleration formula to find the velocity at each point along the circle. 

At the bottom, two forces are applied to the object: the downward force of gravity, $mg$, and the upward tension force. The resultant of these two forces makes a centripetal acceleration toward the center of the circle. 

Applying Newton's second law in this direction (by choosing upward as the positive direction since acceleration at this point is upward), we have \begin{align*} F_c-mg&=\frac{mv^2}{r}\\\\ \Rightarrow F_c &=mg+\frac{mv^2}{r}\\\\&=(515)(10)+\frac{515\times (20)^2}{10}\\\\&=25750\quad {\rm N}\end{align*}

Problem (12): A 2000-kg car is moving around a curve with a radius of 200 m at a speed of 25 m/s. Calculate (a) The centripetal acceleration of the car. (b) The force causing this kind of acceleration. (c) The minimum coefficient of static friction between the tires and the surface of the road guarantees a safe turning.

Solution : The free-body diagram below (side view) shows all forces acting on the car.

(a) Centripetal acceleration is found by the following formula \[a_c=\frac{v^2}{r}=\frac{25^2}{200}=3.125\quad {\rm \frac{m}{s^2}}\] (b) The force along this direction is found using Newton's second law of motion as below \[F_c=ma_c=2000\times 3.125=6250\quad {\rm N}\] (c) When a car turns a curve, it experiences a centripetal acceleration toward the center of the curve. The force causing this acceleration for a car around a curve is static friction force. 

If the static friction force is not great enough, the car will skid out of the curve and follow a nearly straight line. In this situation, the static friction must be balanced with the force needed to produce the given centripetal acceleration. 

On the other hand, recall that the maximum value of static friction force is given by $f_{s, max}=\mu_s N$, where $N$ is the normal force on the car, which is equal to the weight $mg$ (only for flat circular tracks). Thus, \begin{align*} \mu_s (mg)&=\frac{mv^2}{r}\\\\ \Rightarrow \mu_s&=\frac{v^2}{rg}\\\\&=\frac{25^2}{200\times 9.8}\\\\&=0.32\end{align*} Therefore, the car can make a safe round if the coefficient of static friction is greater than this value.

Practice these questions to understand friction force: Problems on the coefficient of friction

Problem (13): A 1500-kg car moves around a flat circular track with a radius of 30 m. The coefficient of friction between the car's tires and the road is 0.3. Find the maximum speed at which the car turns the track.

Solution : The static friction between the car's tires and the road provides the force required for turning the car around the circular track. 

The maximum value of the static friction force formula is $f_{s, max}=\mu_s N$, where $N$ is the force exerted on the car from the surface of the road, which is called the normal force. By setting Newton's second law in the vertical direction to zero (because the car does not fly!), we can find the normal force as below \[N-mg=0 \Rightarrow N=mg\] Next, apply Newton's second law of motion again to the radial direction. 

Because the car moves around a curve, it experiences centripetal acceleration provided by static friction. Thus, \begin{align*} f_{s,max}&=\frac{mv^2}r \\ \\\mu_s (mg)&=\frac{mv^2}r\\\\ \Rightarrow\quad v&=\sqrt{r\mu_s g}\\\\&=\sqrt{30\times (0.3)\times 9.8}\\\\&=9.4\quad {\rm \frac ms}\end{align*}

Problem (14): A car is to turn a curve track with a radius of 120 m at a speed of 85 km/h. How large must the coefficient of static friction be between the car's tire and the road to maintain safe travel?

Solution : The forces on the car are the downward gravity force, $mg$, the normal force $N$ exerted upward by the road, and the static friction force due to the road. All these forces are shown in the free-body diagram below. 

In the vertical direction, there is no acceleration. By applying Newton's second law, we find that the gravity force is equal to the normal force. \[\Sigma F_y=0 \rightarrow N-mg=0 \Rightarrow N=mg\] In the radial direction, the net force required to keep the car on a circular path is found using the centripetal force formula, $F_c=\frac{mv^2}{r}$. 

This force must be balanced with the static friction force to guarantee a safe turn. Thus, the required condition for not skidding on the curve is as follows \[F_c=f_{s,max}\] where $f_{s,max}=\mu_s N$ is the maximum value of the friction force. Therefore, \begin{align*} \frac{mv^2}{r}&=\mu_s (mg) \\\\\Rightarrow \quad \mu_s&=\frac{v^2}{r\,g}\\\\&=\frac{(23.6)^2}{10\times 120}\\\\&=0.4\end{align*}

In the above, we converted ``km/h'' to ``m/s'' as below \[ 85\,{\rm \frac{km}{h}}=85\,\rm{\left(\frac{1000\,m}{3600\,s}\right)}=23.6\,{\rm \frac ms}\]

Problem (15): A 2-kg bucket at the end of a rope is circulated in a vertical plane with a radius of 1.5 m. The tension of the rope at the lowest point of the path is 30 N. Calculate  (a) The speed of the bucket. (b) How fast must the bucket at the top of the path move so that the rope does not go slack?

Solution : The gravity force, $mg$, and the tension in the rope are exerted on the bucket, which is moving in a circular motion. A free-body diagram shows these forces below.

(a) Apply Newton's second law at the lowest point of the path to find the equation governing that point. 

We choose the upward direction as positive because it is in the direction of centripetal acceleration (radially inward). \begin{align*} T-mg&=\frac{mv^2}{r}\\\\ 30-2(10)&=\frac{2\times v^2}{1.5}\\\\ 10\times 1.5&=2v^2\\\\\Rightarrow v&=2.8\,{\rm \frac ms}\end{align*} (b) The sum of the forces at the highest point of the circle, using Newton's second law, is written as below (down is considered as positive) \begin{align*} T+mg&=\frac{mv^2}r\\\\ \Rightarrow v&= \sqrt{\frac{r(T+mg)}{m}}\end{align*} If we set $T=0$ in the above, the rope is on the verge of going slack. Therefore, \[v= \sqrt{\frac{r(T+mg)}{m}}= \sqrt{\frac{(1.5)(0+2\times 10)}{2}}=3.8\,{\rm \frac ms}\]

Problem (16): A jet plane moves around a vertical circular loop. Its speed at the lowest point of the circle is 950 km/h. Find the minimum radius of the circular path so that the (centripetal) acceleration exerted on it does not exceed 5 g's. 

Solution : The jet moves around a circular path and must experience centripetal acceleration. Set this acceleration to $5g$ and solve for $r$. Thus, we have \begin{align*} a_c&=5g \\\\\frac{v^2}{r}&=5g \\\\\Rightarrow r&=\frac{v^2}{5g}\\\\&=\frac{(950\,\frac {km}{h})\left(\frac{1000\,m}{3600\,s}\right)}{5(9.8)}\\\\&=1421\,{\rm m}\end{align*} (b) The weight of the pilot is 75 kg. What is its apparent weight at the bottom and top of the circle?

The apparent weight is equal to the net force on the pilot in the plane. Two forces act on the pilot. One is the downward gravity force, $mg$, and the other is the upward normal force at the lowest point of the circular path.  These two forces acting on the pilot are shown in the free-body diagram below.

The vector sum of these two forces provides the centripetal acceleration around the curved path. \[\Sigma F_r=N-mg=m\frac{v^2}{r}\] where we have chosen up as the direction of the acceleration (radially inward) at this point, as the positive direction. Therefore, \begin{align*} N&=mg+mv^{2}/r\\ &=mg+m(5g)\\&=6mg\\&=6(75)(9.8)\\&=4410\,{\rm N}\end{align*} In all such vertical circular problems, the normal force is the apparent weight. 

Problem (17): The radius of a highway curve banked at an angle of $13^\circ$ is 75 m. At what speed can a car make a turn this curve without the help of friction?

Solution : To solve this problem, we can use the concept of centripetal force. When a car turns on a curved road, the centripetal force required to keep it moving in a circle is provided by the friction between the tires and the road surface.

However, in this case, we are asked to find the speed at which the car can make the turn without relying on friction. This means that there is no horizontal component of friction acting on the car.

Two real forces act on the car: the force of gravity, and the normal force $\vec{N}$. Establish a coordinate system so that the centripetal force lies in the radial direction as depicted in the figure below.  

The vertical component of the normal force provides the necessary centripetal force in this situation. The normal force can be resolved into two components. 

The $y$-component of the normal force cancels out with gravity, while the radial component provides the centripetal force. Write Newton's second law along the $y$-direction: \[N\cos\theta=mg \] On the other hand, the radial component of the normal force, $N\sin\theta$, is the same as the centripetal force, which is given by: \begin{gather*} F_{\text{centripetal}} = N\sin\theta \\\\ \frac{mv^2}{r}=N\sin\theta \end{gather*} Dividing these two last expressions, we get a relation as follows \[\tan\theta=\frac{v^2}{rg}\] Rearranging that, solving for $v$ and substituting the given numerical values into it yields \begin{align*} v&=\sqrt{rg\tan\theta}\\\\ &=\sqrt{75\times 9.8\times \tan 13^\circ} \\\\ &=13.02\,\rm m/s\end{align*} Therefore, the car can make the turn on this curve without the help of friction at a speed of approximately 13 m/s.

Problem (18): An electron is moving in a circular track with a radius of 2 mm at a speed of $2\times 10^{6}\,{\rm m/s}$. Find (a) The period of the revolutions; (b) The centripetal acceleration experienced by the electron.

Solution : (a) This electron, every 1 second covers a distance of $2\times 10^6 \,{\rm m/s}$ which is the circumference of the circle. 

Recall that the circumference of a circle of radius $r$ is $2\pi r$, thus we must first find how many numbers the electron revolves around this circular track. \[L=N(2\pi r) \Rightarrow N=\frac{L}{2\pi}\] On the other hand, the period is the time it takes the electron to make one revolution in one second. According to this definition, the period of an object with $N$ revolutions is obtained as \[T=\frac{1}{N}=\frac{2\pi r}{L}\] Using this formula, we can find the period of this electron \[T=\frac{2\pi\times (0.002)}{2\times 10^6}=6.3\times 10^{-9}\,{\rm s}\] Hence, the period of the electron is about $6\,{\rm ns}$. 

(b) The electron moves along a circular motion whose centripetal acceleration is computed as below \[a_c=\frac{v^2}{r}=\frac{(2\times 10^6)^2}{0.002}=10^9\quad {\rm \frac {m}{s^2}}\]

Problem (19): A 0.45-kg ball is fastened to a rope and is swung in a horizontal plane (without friction) around a circle of radius 1.3 m. When the tension in the rope exceeds 75 N, the rope is broken. What is the maximum speed the ball can have at this point?

Solution: In this circular motion problem, the only force causing the centripetal acceleration is tension in the rope. By applying the centripetal force formula, $F_c=mv^2/r$, and solving for unknown speed $v$, we get \[v=\sqrt{\frac{rF_c}{m}}=\sqrt{\frac{1.3\times 75}{0.45}}=14.7\,{\rm \frac ms}\]

Problem (20): How fast must a centrifuge revolve if a point particle located 5 cm from its axis of rotation is to experience a centripetal acceleration of 100,000 g's?

Solution: The particle in a centrifuge has a circular motion with a constant speed. Solving for $v$ in the centripetal acceleration, $a_c=v^2/r$, we have \[v=\sqrt{ra_c}=\sqrt{0.05\times 100,000}=32\,{\rm \frac ms}\]

Problem (21): What is the acceleration of a stone attached to the end of a 2-m-long rope revolving at 45 revolutions per minute (rpm)? 

Solution : The stone has a circular motion with the given constant speed. In this problem, the speed of a revolving object is given in rpm or angular speed. 

To use the centripetal acceleration formula, $a_c=v^2/r$, we need speed in $m/s$. To convert angular speed (rpm) into linear speed (m/s), we proceed as below formula \[v(m/s)=r\times \frac{rpm \times 2\pi}{60}\] Therefore, an angular speed of "45 rpm" is converted into "m/s" as below \[v=2\times \frac{2\pi \times 45}{60}=9.5 \,{\rm m/s}\] Now, we can plug this value into the centripetal acceleration formula \[a_c=\frac{v^2}{r}=\frac{(9.5)^2}{2}=45.1\quad {\rm \frac {m}{s^2}}\]

Problem (22): A car moves along a curved path as shown in the following figure. When the car reaches the bottom of the track, find (a) the normal force exerted on the car and (b) the effective weight (apparent weight) of the car.

Solution : At the bottom of the track, two forces act on the car. The downward gravity force, $mg$, and the upward force due to the contact between the car and the surface, which is known as the normal force, $N$. 

The car rounds a curve, so it experiences a centripetal force $F_c$ radially inward in the same direction as centripetal acceleration. At the bottom of the track, this force is directed upward.

It is better to choose this direction as positive and apply Newton's second law along this direction as below: \begin{gather*} N-mg=\frac{mv^2}{r} \\\\ \Rightarrow N=mg+\frac{mv^2}{r} \end{gather*}

Problem (23): A centripetal force $F$ is exerted on an object moving around a circular track with a constant speed $v$. If the speed of the object is tripled and the radius of the track is quadrupled, what happens to the centripetal force? 

Solution: The centripetal force in a circular motion is given by the equation $F_c=\frac{mv^2}{r}$. As you can see, this kind of force is proportional directly to the speed of the object and inversely to the radius of the curved path. Therefore, \begin{align*} \frac{F_2}{F_1}&=\left(\frac{v_2}{v_1}\right)^2 \frac{r_1}{r_2}\\\\&=\left(\frac{3v_1}{v_1}\right)^2 \frac{r_1}{4r_1}\\\\&=\frac{9}{4}\end{align*} Hence, the force becomes 2.25 times the original force. \[F_2=\frac 94 F_1\]

Problem (24): A stone is attached to the end of a rope of length $L$ and rotates in a vertical circle. At the top of the path, the tension in the rope is twice the weight of the stone.  (a) Determine the net force on the stone when it reaches the highest point of the path. (b) What is the speed of the stone at this point? After a few circles, when the stone is at the top of the circle, the rope is broken. (c) Determine the time it takes the stone to hit the ground.  (d) Find the horizontal distance traveled by the stone before striking the ground.

Solution:  This is left up to you to solve.

In this article, we learned how to solve problems involving circular motion. The most important notes we practiced were as follows:

•  An object moving in a circular path of radius $r$ at constant (uniform) speed $v$ experiences an acceleration whose magnitude is $a_c=\frac{v^2}{r}=r\omega^2$ and its direction is toward the center of the circle. 
• Uniform circular motion is an example that shows that acceleration and velocity are not always in the same direction. 
• In such motions, $\vec{a}$ and $\vec{v}$ are perpendicular. 

Author : Dr. Ali Nemati Date Published:  7-31-2021  

© 2015 All rights reserved. by Physexams.com

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Classical Mechanics

Herbert goldstein, charles p. poole jr., john l. safko, the central force problem - all with video answers.

Chapter Questions

Consider a system in which the total forces acting on the particles consist of conservative forces $\mathbf{F}_{i}^{\prime}$ and frictional forces $\mathbf{f}_{i}$ proportional to the velocity. Show that for such a system the virial theorem holds in the form $$\bar{T}--\frac{1}{2} \sum \mathbf{F}_{i}^{\prime} \cdot \mathbf{r}_{i}$$ providing the motion reaches a steady state and is not allowed to die down as a result of the frictional forces.

By expanding $c \sin \psi$ in a Four series in at, show that Kepler's equation has the formal solution $$y^{\prime}=\operatorname{tat}+\sum_{x=1}^{\infty} \frac{2}{n} J_{n}(n c) \sin \omega t$$ Where $J_{n}$ is the Bessel function of order $n$. For small argument, the Bessel function can be approximated in a power series of the argument. Accordingly, from this result derive the first few Terms in the expansion of $\psi$ in power of $e$.

If the difference $y$ - ar is represented by $p$, Kepler's creation can be written $$\rho-c \sin (\omega t+\rho)$$. Successive approximations to $\rho$ can be obtained by expatiating sin $\rho$ in a Taylor vines In $\rho,$ and then replacing $\rho$ by is expression given by Kepler's equation. Show that the first approximation by $\rho$ is $\rho_{1},$ given by $$\tan \rho_{1}-\frac{c \sin c t}{1-e \cos \theta d}$$, and that the next approximation is found from \[ \sin \left(p_{2}-p_{1}\right)--c^{3} \sin \left(\cot +p_{1}\right)(1+\epsilon \cos (u l) \] an expression that is accurate through terms of order $c^{4}$.

Show that for repulsive scattering. Eq (3.96) for the angle of watering as a function of the impact parameter. $x$, can written as $$\Theta=\pi-4 s \int_{0}^{\mathrm{t}} \frac{\rho d \rho}{\sqrt{r_{\mathrm{rn}}^{2}\left(1-\frac{v}{\mathrm{E}}\right)-\mathrm{s}^{2}\left(1-\rho^{2}\right)}}$$, Derivations or $$\theta=\pi-4 s \int_{0}^{1} \frac{d \rho}{\sqrt{\frac{r^{2}}{\rho^{2} E}\left(V\left(r_{m}\right)-V(r)\right)+s^{2}\left(2-\rho^{2}\right)}}$$, by changing the variable of integration to some function $\rho(r)$. Show that for a re pulsive potential the integrated is ncyer singular in the limit $r \rightarrow r_{m}$. Because of the definite limits of integration, these formulations have advantages for numerical calculations of $\Theta(x)$ and allow naturally for the tere of Gaths-legendre quadrature schemes

Apply the formulation of the preceding exercise to compute numerically $\Theta(s)$ and the differential cross section of $\sigma(\Theta)$ for the repulsive potential $$v=\frac{v_{0}}{1+r}$$ and for a total energy $E-1,2 \mathrm{V}_{0}$. It is suggested that 16 -point Gauss Legendre quadrature will give loquacity accuracy. Does the scatting inhibit a rainbow?

If a repulsive potential drops of monotonically with $r,$ then for energies high compared to $V\left(r_{m}\right)$ the angle of scattering will he small. Under these conditions show that Eq. (3.97) can be manipulated so that the deflection angle is given approximately by $$\Theta=\frac{1}{\varepsilon} \int_{0}^{1} \frac{\left(V\left(u_{m}\right)-V(u)\right) d y}{\left(1-y^{2}\right)^{3 / 2}}$$, where $y$, otwicusly, is $u / t u p$ Show further, that if $V(u)$ is of the form $C u^{n}$, where $n$ is a positive integer, then in the high energy limit the cross section is proportional to $\theta^{-2(1+1 / n)}$.

(a) Show that the angle of recoil of the target particle relative to the incident direction of the scattered particle is simply $\Phi=\frac{1}{2}(\pi-\Theta)$ (b) It is obscured that in classic scattering the scattering cross section is the in Terms of $\Theta$. What are the corresponding probability distributions for the scaftered energy of the incidcat particle, $E_{1}$, and for the recoil energy of the target particle. $E_{2} ?$

Show that the angle of scattering in the laboratory system, $\theta,$ b related to the cretgy before scattering. $E_{q}$ and the energy after scattering $E_{1}$. according to the actuation $$\cos \theta=\frac{m_{2}+m_{1}}{2 m_{1}} \sqrt{\frac{E_{1}}{E_{0}}} \quad \frac{m_{2}-m_{1}}{2 m_{1}} \sqrt{\frac{E_{0}}{E_{1}}}+\frac{m_{2} Q}{2 m_{1} \sqrt{E_{0} E_{1}}}$$.

Show that the central force problem is sclutile in terms of cliptic functions when the force is a power-law function of the distance with the following fractional exponents: \[ n=-\frac{3}{2},-\frac{5}{2},-\frac{1}{3},-\frac{5}{3},-\frac{7}{3} \] $$n=-\frac{3}{2},-\frac{5}{2},-\frac{1}{3},-\frac{5}{3},-\frac{7}{3}$$

A planet of mass $M f$ is in an torture of eccentricity $e=1-a$ where $a<1$, about the Sivan. Assume the thrion of the Sun can be neglected and that only gravitational forces ack. When the planel is at its ereatest distance from the Sun, it is struck by a cornet of mass $m$. Where $m<M$ traveling in a tangenting direction. Assuming the collision is completely inelastic, find the minionum kinetic energy the conet mast have to change the new ortoit to a parabola.

Two particles move about each other in circular orbits under the influence of gravitational forces, with a period $\tau$. Their mourn is suddenly stopped a a given instant of time. and they are then released and allowed to fall into each other. Prove that they collide after a time $r / 4 \sqrt{2}$.

Suppose that there are long-range interactions between atoms in a gas in the form of centril forces derivable from a potential $$U(r)-\frac{k}{r^{m}}$$, where $r$ is the distance between any pair of utcems and $n t$ is is positive integez. Assume further that relative to any your atom the other atoms inc distributed in spice such that the volume density is yiven hy the Bolicmenn farior: \[ \rho(r)=\frac{N}{V} e^{-t / \varphi \backslash / R T} \] where $N$ is the total number of worms in a volume $V$. Find the condition to the virial of Claustus resulting from the forces between pains of aforns. and compute the resulting currection to Boyic's law. Take $N$ so large that surns may be replaced by interlinks While closed results can be found for any positive $m$, if desired, the mathematics can be simplified by taking $m-+1$.

(a) Show that if a particle describes a circular orbit under the influence of an attractive central force directed toward a point on the circle. then the force varies as the imeric-fint power of the distance. (b) Show that for the orbit described the total energy of the parbele is zero. (c) Find the period of the motion. (d) Find $x_{0}, y_{4}$ and $v$ as a function of angle around the circle and bow that all three quarititics aro intinite as the particle goes through the center of force.

(a) For circular and parabolic orbits in an attractive $1 / r$ potential having the same insular momentum. show that the perihelion distance of the parabola is one-half the radius of the circle (b) Prove that in the same central force as in pirt (a) the speed of a particle at are point in a parabolic orbit is $\sqrt{2}$ times the sped in a circular orbit passing through the same point.

A meter is observed to stribe Earth with a speed $v$. making an angle $\phi$ with the acnith. Suppose that fas trom Earth the meter's speed was $s^{\prime}$ and it was persceeting in a direction making a renith angle $\phi$, the effect of Earth's gravity being to pull it into a byperbolic orbit intersecting Earth's surface. Show how $v$ and $\phi^{\prime}$ can be determined from $w$ und $\phi$ in terme of lanown constants.

Prove that in a Kepler elliptic orbit with small eccentricity $e$ the ungular motion of a particle as viewed from the cmpry focus of the ellipse is uniform (the cmpty focus is the focus that is aot the center of attraction) to first order in $e$. It is this theorem thit crables the Ptolemaic picture of plarctary motion to be a reasonably accurate approximation. On this picture the Sun is resumed to move uniformly on a circle taken as the distance between the two foci of the connect elliptical catur, then the angular motion is thus described by the Ptolemaic picture accurately to tirst onder in $\boldsymbol{e$)

One classic there in science fiction is a twin plaret ("Planet $X$ ) to Earth that is identical in mass, energy, and Mentholatum twit is located on the urtut $90^{\circ}$ cut of phase with Earth so that it is lidulca froce the Sun. However, because of the elliptical nature of the critic it is not always completely hidden. Assume this twin planet is in the same Keplenan ortut as Earth in such a manner the is ts in apheling when Earth is in peribetion. Calculate to hist order in the egocentricity $e$ the maximentn angular separation of the twin and the Sun as viewed from the Earth. Could such a twin be visible from Earth? Suppose the twin planet is in an elliptical orbit having the seem sice and shope as that of Earth, but rotated $180^{\circ}$ from Earth's orbit, so that Earth and the twin are in rebellion at the same time. Repeat your calculation and compare the visibility in the two situations.

At perimeter of an elliptic motivational croft a particle experiences an impulse $S$ (c). Exercise 11 . Chapter 2 ) in the radial correction, chiding the Pericles into another elliptic orbit. Determine the new semimonthly axis, eccentricity, and orientation in terms of the old.

A particle mesas in a force fared described by \[ F(r)=\frac{k}{r^{2}} \exp \left(-\frac{r}{a}\right) \] where $k$ and $a$ are positive. (a) Write the equations of motion and reduce them to the equivalent cons-dimensional Problem Use the effectroe potenthal to discuss the qualitative attire of the orturs for different values uf the energy and the angular momentous. (b) Shose that if the crbut is nearly circular, the upsides will advance approximately by $\pi \rho / a$ per revolution. where $\rho$ is the radius of the circular orbit.

A uniform distribution of dust in the solar system adds to the gravitational attraction of the Sun on a planet an additional force \[ \mathbf{F}=m C \mathbf{r} \] where $m$ is the mass of the planet, $C$ is a constant proportional to the gravitational constant and the density of the dust, and $\mathbf{r}$ is the radius vector from the Sum to the Planet (both considered as paints). The additional force is very small compared to the direct Sun-planet gravitational force. (a) Calculate the period for a circular orbit of radius $r_{0}$ of the planet in this combined field. (b) Calculate the period of radial oscillations for slight disturbances from this circular orbit. (c) Show that nearly circular orbits can be approximation by a processing eclipse and find the procession frequency. Is the procession in the same or apposite dialectical In the critic angular velocity?

Show that the motion of a particle in the potential field $$V(r)--\frac{k}{r}+\frac{h}{r^{2}}$$ is the same as the of the osmotic under the Kepier potential alone when crevessed in tempts of a coordinate system rotating processioning around the center of force. For negative total energy, show that if the additional potential term is very small compared to the Kepier potential. then the angular speed of precessica of the catalytic orbit is $$\vec{\Omega}=\frac{2 \pi m h}{l^{2} r}$$. The perihelion of Mencury is chserved to precess (after correction for known planctary pcrurtutions) at the rate of about $40^{\circ}$ of are per century. Show that this precessica could be accounted for clascicaliy if the dimensionless quantiy \[ \eta-\frac{h}{k a} \] (which is a then sure of the perturbing incentive-guarder potential relative to the gravitational political were as scroll $x=10^{-8}$. (The egocentricity of Mercury's orbit is $0.206,$ and its period is 0.24 year.)

The additional term in the potential behaving as $r^{-2}$ in Exercise 21 looks very mach like the centrifugal berries term in the equivalent one-dimensional potential. Why is it then fast the additional force term causes a precession of the orbit. while an addition to the turner. through a change in $l$, does not?

Evaluate approximately the ratio of mass of the Sun to that of Earth, aching only the Lengths of the year and of the lunar month $(27.3$ divy), and the mean radio of Earth's tortoise $\left(1.49 \times 10^{8} \mathrm{km}\right)$ and of the Moon's orbicular $\left(3.8 \times 10^{5} \mathrm{km}\right)$.

Show that for elliptical motion in a gravitational field the radial speed can be written as \[ r=\frac{\tan }{r} \sqrt{a^{2} c^{2}-(r-a)^{2}} \] Introduce the eccentric anomaly variable $\psi$ in place of $r$ and show that the resulting differential equation in $\forall$ can be integrated immediately to give Kepler's actuation.

If the eccentricity $c$ is small, Kepler's equation for the concoction anrenaly $\psi$ as a function of art$,$ Eq $,(3.76)$, is easily solved on a computer by an itcative techalque that trats the $c \sin \psi$ term as of lower order than $\downarrow$. Denoting $\forall$ by the nith lierative titivation, the obvious iteration relation is \[ \psi_{n}=a r+c \sin \psi_{n-1} \] Using this iteration procedure, firs the analytic firm: for an expansion of $\forall$ in powers of $c$ at least through terms in $t^{3}$

Earth's period between successive perception transits (the "normalize year") in 365.2596 mean solar days, and the eccentricity of its orbit is 0.0167504 . Assuming motion in a Keplerian elliptical orbit, bow far does the Earth move in angle in the orbit, starting from perihelion, in a time equal to one quarter of the anomalistic year? Give your result in degrees to an accuracy of one second of are ec better Any method may be used, including numerical computation with a calculator or computer.

In hyperbolic motion in a $1 / r$ potential, the analogy of the vertices anomaly is $F$ defined by $$r=a(e \cosh F-1)$$. where $a(e-1)$ is the distance of closest approach. Find the analogue to Kepler's equation giving $t$ from the time of closest approach ss a function of $\bar{F}$.

A magnetic monopole is defined (if one exists) by a magnetic field singularity of the form $\mathbf{B}-$ br $/ r$, where $b$ is a constant ta measure of the magnetic charge, is it were) Suppose a particle of mass $m$ moves in the field of a mingnctac monopolize and a central force field derived from the potential $V(r)--k / r$. (a) Find the form of Ncwicris equation of motion, usirg the Lorectz force given by Eq. $(1.60) .$ By lookits at the product $r \times p$ show that while the mechanical angular mornchturn is are conserved (the field of force is noncentral) there is a conserved vecior \[ \mathbf{D}=\mathbf{L}-\frac{q b \mathbf{r}}{c} \] (b) By paralleling the steps leading from Eq (3,79) to Eq. $(3 \mathrm{S} 2)$. show that for some $f(r)$ thete is a construed vector uraloscus to the Leplace- Runge Lenz vector in which $\mathbf{D}$ plays the same role as $\mathbf{L}$, in the pure Kepler force problem.

If a] the momentum vectors of a particle along its trajectory are translated so as to stant from the center of force, then the hendw of the vectons trece out the particle's hodograyh, a locus curve of considerable artiçuity in the history of mechanics. with something of a revival in comncection with space vehicle dynamica, By taking the croses product of $\mathbf{L}$. Writh the Laplace- Runge Lenz vector A. show that the hadograph for cliiptical Keplet motion ts a circle of tadius $m k / l$ with origin on the $y$ aris displaced a distance $A / l$ from the conter of force.

What charges, if any, would there be in Rutherford scattering if the Coulomb force were attractive, instead of repulsive?

Examine the criticizing produced by a repulsive central force $f=k r^{-3}$. Show that the differential cross section is given by \[ \sigma(\theta) d \theta-\frac{A}{2 E} \frac{(1-x) d x}{x^{2}(2 \quad x)^{2} \sin \pi x} \] where $x$ is the ratio of $\Theta / x$ asd $E$ is the energy.

A contral forte potentid frequently encountered in nucleur physics is the rectungular well. defined by the potcatial \[ \begin{aligned} V &=0 & & r>a \\ &=-V_{0} & & r \leq a \end{aligned} \] Show that the scatting produced by such a potcrital in classical bicchanics is identical with the refraction of light rays by a sphere of radius $a$ and relative index of nefraction \[ n=\sqrt{\frac{E+V_{0}}{E}} \] Mhis equinalence domonstrates why it was pessible to explain refraction phenoment hoch by Huypen's waves and by Newton's machanical corpuscles.) Show also thin the differcritial cross section is \[ \sigma(\theta)=\frac{n^{2} \alpha^{2}}{4 \cos \frac{\theta}{2}} \frac{\left(n \cos \frac{\theta}{2}-1\right)\left(n-\cos \frac{\theta}{2}\right)}{\left(1+n^{2}-2 n \cos \frac{t}{2}\right)^{2}} \] What is the total cross section?

A particle of mass $m$ is constrained to move under gravity without friction on the inside of a parabolaid of revolution whose axis is vertical. Find the cone-dimensional problem equivalent to its motion. What is the cons ticu ca the particle's initial velocity to produce circular motion? Find the period of small oscilletions about this circular motion.

Consider a truncated repulcive Coulomb potential definod as \[ \begin{aligned} V &-\frac{k}{r} \\ &=\frac{k}{a} \quad r \leq a \end{aligned} \] For a particle of total energy $E>k / a$. obtivin expressions for the scamering angic $\Theta$ as a function of $s / \mathrm{s}_{0}$. Where so is the impuat parameter for which the periapis occurs deduce about the angular scattering cross section from the dependence of $\theta$ on $s / 20$ for this particular case?

Another vervion of the truncated Coulomb percatial has the form \[ \begin{aligned} V &-\frac{k}{r}-\frac{k}{a} & & r>a \\ &=0 & & r &<a \end{aligned} \] Obtain closed form expressions for the scattering angle and the differential scattering cross section. These are moel cursenterity expressed in terms of a parameter rucasartng the distance of closest wprouch in units of a. What is the totel cross section?

The restricted three-hody problem consists of two masses in circular crhits about each other and a third body of thuch smaller mass whose cffect on the twe larger hodies can be neglected. (a) Define an effective putcrial $V(x, y)$ for this problems where the $x$ axis is the line of the two laner masses. Sketch the function $V(x, 0)$ and show that there are two "valleys" ipcints of stuble cquiliticium) corresponding to the two masses. Also shew that there are three "hills" (three peints of unstable equilihrium). (b) Lising a computer program, calculate scuse orbits for the restricted three body problem. Maty orbits will end with cjcction of the smaller mass. Start by assum the a position and a vector velocity for the small mass,

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Simplifying a Difficult Problem

Consider the situation below in which a force is directed at an angle to the horizontal. In such a situation, the applied force could be resolved into two components. These two components can be considered to replace the applied force at an angle. By doing so, the situation simplifies into a familiar situation in which all the forces are directed horizontally and vertically.

Once the situation has been simplified, the problem can be solved like any other problem. The task of determining the acceleration involves first determining the net force by adding up all the forces as vectors and then dividing the net force by the mass to determine the acceleration. In the above situation, the vertical forces are balanced (i.e., F grav , F y , and F norm add up to 0 N), and the horizontal forces add up to 29.3 N, right (i.e., 69.3 N, right + 40 N, left = 29.3 N, right). The net force is 29.3 N, right and the mass is 10 kg (m = F grav /g); therefore, the acceleration is 2.93 m/s/s, right.

Your Turn to Practice

To test your understanding, analyze the two situations below to determine the net force and the acceleration. When finished, click the button to view the answers.

The net force is 69.9 N, right and the acceleration is 3.5 m/s/s, right .

Note that the vertical forces balance but the horizontal forces do not. The net force is

F net = 129.9 N, right - 60 N, left = 69.9 N, right

The mass is

m = (F grav / g) = 20 kg

So the acceleration is

a = (69.9 N) / (20 kg) =3.50 m/s/s.

The net force is 30.7 N, right and the acceleration is 1.23 m/s/s, right .

F net = 70.7 N, right - 40 N, left = 30.7 N, right

m = (F grav / g) = 25 kg

a = (30.7 N) / (25 kg) =1.23 m/s/s.

What's Up with the Normal Force?

There is one peculiarity about these types of problems that you need to be aware of. The normal force (F norm ) is not necessarily equal to the gravitational force (F grav ) as it has been in problems that we have previously seen. The principle is that the vertical forces must balance if there is no vertical acceleration. If an object is being dragged across a horizontal surface, then there is no vertical acceleration. For this reason, the normal force (F norm ) plus the vertical component (F y ) of the applied force must balance the gravitational force (F grav ). A quick review of these problems shows that this is the case. If there is an acceleration for an object being pulled across a floor, then it is a horizontal acceleration; and thus the only imbalance of force would be in the horizontal direction .

Now consider the following situation in which a force analysis must be conducted to fill in all the blanks and to determine the net force and acceleration. In a case such as this, a thorough understanding of the relationships between the various quantities must be fully understood. Make an effort to solve this problem. When finished, click the button to view the answers. (When you run into difficulties, consult the help from a previous unit .)

 The F grav is

F grav = m • g = (10 kg) • (9.8 m/s/s) = 98 N

Using the sine function,

F y = (60 N) • sine (30 degrees) = 30 N

Since vertical forces are balanced, F norm = 68 N.

Now F frict can be found

F frict = mu • F norm = ( 0.3) • (68 N) = 20.4 N

Using the cosine function,

F x = (60 N) • cosine (30 degrees) = 52.0 N

Now since all the individual force values are known, the F net can be found:

F net = 52.0 N,right + 20.4 N, left = 31.6 N, right.

The acceleration is

a = 3.16 m/s/ s, right.

In conclusion, a situation involving a force at an angle can be simplified by using trigonometric relations to resolve that force into two components. Such a situation can be analyzed like any other situation involving individual forces. The net force can be determined by adding all the forces as vectors and the acceleration can be determined as the ratio of Fnet/mass.

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Check your understanding.

The following problems provide plenty of practice with F net = m • a problems involving forces at angles. Try each problem and then click the button to view the answers.

Glen Brook and Warren Peace are incorrect. Warren Peace perhaps believes that the Fnorm = Fgrav; but this is only the case when there are only two vertical forces and no vertical acceleration; sorry Warren - there is a second vertical force in this problem (F app ).

Glen Brook perhaps thinks that the F app force is 50 N upwards and thus the F norm must be 50 N upwards to balance the Fgrav. Sorry Glen - the F app is only 25 N upwards (50 N) • sine 30 degrees).

"Datagal Olive!"

2. A box is pulled at a constant speed of 0.40 m/s across a frictional surface. Perform an extensive analysis of the diagram below to determine the values for the blanks.

First use the mass to determine the force of gravity.

F grav = m • g = (20 kg) • (9.8 m/s/s) = 196 N

Now find the vertical component of the applied force using a trigonometric function.

F y = (80 N) • sine (45 degrees) = 56.7 N

Thus, F norm =139.3 N in order for the vertical forces to balance.

The horizontal component of the applied force can be found as

F x = (80 N) • cosine (45 degrees) = 56.7 N

Since the speed is constant, the horizontal forces must also balance; and so F frict = 56.7 N.

The value of "mu" can be found using the equation

3. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Answer: "mu" = 0.25

 The F grav can be calculated from the mass of the object.

The vertical component of the applied force can be calculated using a trigonometric function:

F y = (80 N) • sine (30 degrees) = 40 N

In order for the vertical forces to balance, F norm + F y = F grav . Thus,

F norm = F grav - F y = = 196 N - 40 N = 156 N

The horizontal component of the applied force can be calculated using a trigonometric function:

F x = (80 N) • cosine (30 degrees) = 69.2 N

The net force is the sum of all the forces when added as vectors. Thus,

F net = (69.2 N, right) + (40 N,left) = 29.2 N, right

a = F net / m = (29.2 N, right) / (20 kg) = 1.46 m/s/s, right.

The value of "mu" can be found using the equation "mu" = F frict / F norm .

4. The 5-kg mass below is moving with a constant speed of 4 m/s to the right. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The F grav can be calculated from the mass of the object.

F grav = m • g = (5 kg) • (9.8 m/s/s) = 49 N

The vertical component of the applied force can be calculated using a trigonometric function.

F y = (15 N) • sine (45 degrees) =10.6 N

F norm = F grav - F y = = 49 N - 10.6 N = 38.4 N

F x = (15 N) • cosine (45 degrees) = 10.6 N

Since the speed is constant, the horizontal forces must balance. Therefore, F frict = 10.6 N.

The value of "mu" can be found using the equation "mu"= F frict / F norm :

"mu" = 0.276

5. The following object is being pulled at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

Since the velocity is constant, the acceleration and the net force are 0 m/s/s and 0 N respectively. The F grav can be calculated from the mass of the object.

The object moves at constant speed; thus, the horizontal forces must balance. For this reason, F x = 10 N.

The applied force can now be found using a trigonometric function and the horizontal component:

cosine (60 degrees) = (10 N) / (F app )

Proper algebra yields

F app = (10 N) / [cosine (60 degrees) ] = 20 N

F y = (20 N) • sine (60 degrees) = 17.3 N

F norm = F grav - F y = 49 N - 17.3 N = 31.7 N

6. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +2 m/s/s.

The net force can be calculated as:

F net = m • a = (10 kg) • (2 m/s/s) = 20 N, right.

F x = (70 N) • cosine (45 degrees) = 49.5 N

The net force is the vector sum of all the forces. Thus, F net = F x + F frict . That is,

20 N, right = 49.5 N, right + F frict

Therefore, F frict must be 29.5 N, left.

F y = (70 N) • sine (45 degrees) = 49.5 N

F norm = F grav - F y = = 98 N - 49.5 N = 48.5 N

The value of "mu" can be found using the equation "mu" = F frict / F norm :

7. Study the diagram below and determine the acceleration of the box and its velocity after being pulled by the applied force for 2.0 seconds.

F y = (50 N) • sine (45 degrees) = 35.4 N

In order for the vertical forces to balance, F norm + F y = F grav . Algebraic rearrangement leads to:

F norm = F grav - F y = 196 N - 35.4 N = 160.6 N

F x = (50 N) • cosine (45 degrees) = 35.4 N

The F net is 35.4 N, right since the only force which is not balanced is F x .

The acceleration is:

a = F net / m = (35.4 N) / (20 kg) = 1.77 m/s/s

The velocity after 2.0 seconds can be calculated using a kinematic equation:

v f = v i + a • t = 0 m/s + (1.77 m/s/s) • (2.0 s) v f = 3.54 m/s

8. A student pulls a 2-kg backpack across the ice (assume friction-free) by pulling at a 30-degree angle to the horizontal. The velocity-time graph for the motion is shown. Perform a careful analysis of the situation and determine the applied force.

The slope of a velocity-time graph is the acceleration of the object. In this case, a = +0.125 m/s/s.

F net = m • a = (2 kg) • (0.125 m/s/s) = 0.250 N, right

Since the acceleration is horizontal, the vertical forces balance each other. The horizontal component of the applied force (F x ) supplies the horizontal force required for the acceleration. Thus, the horizontal component of the applied force is 0.250 N.

Using trigonometry, the applied force (F app ) can be calculated:

cosine (30 degrees) = ( 0.250 N) / (F app )

Algebraic rearrangement of this equation leads to:

F app = 0.289 N

9. The following object is moving to the right and encountering the following forces. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F grav = m • g = (10 kg) • (10 m/s/s) = 100 N

F norm = F grav - F y = 100 N - 35.4 N = 64.6 N

This x-component of the applied force (F x ) is directed leftward. This horizontal component of force is not counteracted by a rightward force. For this reason, the net force is 35.4 N.  Knowing F net , allows us to determine the acceleration of the object:

a = F net / m = (35.4 N) / (10 kg) = ~3.5 m/s/s

The acceleration of an object is the velocity change per time. For an acceleration of 3.5 m/s/s, the velocity change should be 3.5 m/s for each second of time change. In the velocity-time table, the velocity is decreasing by 3.5 m/s each second. Thus, the values should read 17.5 m/s, 14.0 m/s, 10.5 m/s, 7.0 m/s, 3.5 m/s, 0 m/s.

10. The 10-kg object is being pulled to the left at a constant speed of 2.5 m/s. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = F grav - F norm = 98 N - 80 N = 18 N

The applied force can be determined using a trigonometric function:

sine 30 (degrees) = (18 N) / F app

Algebraic rearrangement leads to:

F app = (18 N) / [ sine (30 degrees) ] = 36 N

Similar trigonometry allows one to determine the x-component of the applied force:

F x = (36 N) • cosine(30 degrees) = 31.2 N

Since the speed is constant, the horizontal forces must also balance. Thus the force of friction is equal to the F x value. F frict = 31.2 N

The value of "mu" can be found using the equation "mu" = F frict / F norm

11. Use your understanding of force relationships and vector components to fill in the blanks in the following diagram and to determine the net force and acceleration of the object. (F net = m•a; F frict = μ•F norm ; F grav = m•g)

F y = (100 N) • sine (45 degrees) = 70.7 N

F norm = F grav - F y = 98 N - 70.7 N = 27.3 N

F x = (100 N) • cosine (45 degrees) = 70. N

The net force is the vector sum of all the forces.

F net = 50 N, right + 70.7 N, left = 20.7 N, left

The acceleration is can be found from a = F net / m :