central net force particle model circular motion problem solving

6.2 Uniform Circular Motion

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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Classical Mechanics pp 111–152 Cite as

Central Force Problems

• Matthew J. Benacquista 7 &
• Joseph D. Romano 7  
• First Online: 28 February 2018

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Part of the book series: Undergraduate Lecture Notes in Physics ((ULNP))

One of the most ubiquitous forces in physics is the central force. In any case where a particle possesses a “charge” that couples to a “field”, the field outside the charge is nearly always a radial field if the particle can be considered to be a dimensionless point. Examples include the classical Newtonian law of gravitation, the electrostatic field, and the Yukawa model of the strong nuclear force. The common feature in these examples is a spherically-symmetric potential that depends only on the radial separation from the point-particle source of the field. In this chapter, we will focus on mostly on gravitationally-bound systems.

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Alternatively, we can deduce that \(\varvec{\ell }\) is constant by simply noting that for a central potential U ( r ), the force \(\mathbf {F} = -\mathbf {\nabla }U\) points the radial direction, and hence the torque on the particle about the origin is zero. (See Sect.  1.4 , conservation law II.).

Since there are no external forces acting on the system, the center-of-mass frame for this problem is an inertial frame.

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Benacquista, M.J., Romano, J.D. (2018). Central Force Problems. In: Classical Mechanics. Undergraduate Lecture Notes in Physics. Springer, Cham. https://doi.org/10.1007/978-3-319-68780-3_4

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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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6.6: Non-circular Central Motion

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• Page ID 24734

• Peter Dourmashkin
• Massachusetts Institute of Technology via MIT OpenCourseWare

Let’s now consider central motion in a plane that is non-circular. In Figure 6.10, we show the spiral motion of a moving particle. In polar coordinates, the key point is that the time derivative dr / dt of the position function r is no longer zero. The second derivative \(d^{2} r / d t^{2}\) also may or may not be zero. In the following calculation we will drop all explicit references to the time dependence of the various quantities. The position vector is still given by Equation (6.2.1), which we shall repeat below

\[\overrightarrow{\mathbf{r}}=r \hat{\mathbf{r}} \nonumber \]

Because \(d r / d t \neq 0\) when we differentiate Equation (6.5.9), we need to use the product rule

\[\overrightarrow{\mathbf{v}}=\frac{d \overrightarrow{\mathbf{r}}}{d t}=\frac{d r}{d t} \hat{\mathbf{r}}+r \frac{d \hat{\mathbf{r}}}{d t} \nonumber \]

Substituting Equation (6.2.4) into Equation (6.5.10)

\[\overrightarrow{\mathbf{v}}=\frac{d \overrightarrow{\mathbf{r}}}{d t}=\frac{d r}{d t} \hat{\mathbf{r}}+r \frac{d \theta}{d t} \hat{\boldsymbol{\theta}}=v_{r} \hat{\mathbf{r}}+v_{\theta} \hat{\boldsymbol{\theta}} \nonumber \]

The velocity is no longer tangential but now has a radial component as well

\[v_{r}=\frac{d r}{d t} \nonumber \]

In order to determine the acceleration, we now differentiate Equation (6.5.11), again using the product rule, which is now a little more involved:

\[\overrightarrow{\mathbf{a}}=\frac{d \overrightarrow{\mathbf{v}}}{d t}=\frac{d^{2} r}{d t^{2}} \hat{\mathbf{r}}+\frac{d r}{d t} \frac{d \hat{\mathbf{r}}}{d t}+\frac{d r}{d t} \frac{d \theta}{d t} \hat{\boldsymbol{\theta}}+r \frac{d^{2} \theta}{d t^{2}} \hat{\boldsymbol{\theta}}+r \frac{d \theta}{d t} \frac{d \hat{\boldsymbol{\theta}}}{d t} \nonumber \]

Now substitute Equations (6.2.4) and (6.2.7) for the time derivatives of the unit vectors in Equation (6.5.13), and after collecting terms yields

\[\begin{array}{l} \overrightarrow{\mathbf{a}}=\left(\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2}\right) \hat{\mathbf{r}}+\left(2 \frac{d r}{d t} \frac{d \theta}{d t}+r \frac{d^{2} \theta}{d t^{2}}\right) \hat{\boldsymbol{\theta}} \\ =a_{r} \hat{\mathbf{r}}+a_{\theta} \hat{\boldsymbol{\theta}} \end{array} \nonumber \]

The radial and tangential components of the acceleration are now more complicated than then in the case of circular motion due to the non-zero derivatives of \(d r / d t\) and \(d^{2} r / d t^{2}\). The radial component is

\[a_{r}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2} \nonumber \]

and the tangential component is

\[a_{\theta}=2 \frac{d r}{d t} \frac{d \theta}{d t}+r \frac{d^{2} \theta}{d t^{2}} \nonumber \]

The firs term in the tangential component of the acceleration, \(2(d r / d t)(d \theta / d t)\) has a special name, the coriolis acceleration,

\[a_{c o r}=2 \frac{d r}{d t} \frac{d \theta}{d t} \nonumber \]

Example 6.4 Spiral Motion

A particle moves outward along a spiral starting from the origin at t = 0 . Its trajectory is given by \(r=b \theta\) where b is a positive constant with units \(\left[\mathrm{m} \cdot \mathrm{rad}^{-1}\right] \cdot \theta\) increases in time according to \(\theta=c t^{2}\), where \(c>0\) is a positive constant (with units \(\left[\mathrm{rad} \cdot \mathrm{s}^{-2}\right]\))

a) Determine the acceleration as a function of time.

b) Determine the time at which the radial acceleration is zero.

c) What is the angle when the radial acceleration is zero?

d) Determine the time at which the radial and tangential accelerations have equal magnitude.

a) The position coordinate as a function of time is given by \(r=b \theta=b c t^{2}\). The acceleration is given by Equation (6.5.14). In order to calculate the acceleration, we need to calculate the four derivatives \(d r / d t=2 b c t, d^{2} r / d t^{2}=2 b c, d \theta / d t=2 c t\), and \(d^{2} \theta / d t^{2}=2 c\). The acceleration is then

\[\overrightarrow{\mathbf{a}}=\left(2 b c-4 b c^{3} t^{4}\right) \hat{\mathbf{r}}+\left(8 b c^{2} t^{2}+2 b c^{2} t^{2}\right) \hat{\boldsymbol{\theta}}=\left(2 b c-4 b c^{3} t^{4}\right) \hat{\mathbf{r}}+10 b c^{2} t^{2} \hat{\boldsymbol{\theta}} \nonumber \]

b) The radial acceleration is zero when

\[t_{1}=\left(\frac{1}{2 c^{2}}\right)^{1 / 4} \nonumber \]

c) The angle when the radial acceleration is zero is

\[\theta_{1}=c t_{1}^{2}=\sqrt{2} / 2 \nonumber \]

d) The radial and tangential accelerations have equal magnitude when after some algebra

\[\left(2 b c-4 b c^{3} t^{4}\right)=10 b c^{2} t^{2} \Rightarrow 0=t^{4}+(5 / 2 c) t^{2}-\left(1 / 2 c^{2}\right) \nonumber \]

This equation has as only positive solution for \(t^{2}\)

\[t_{2}^{2}=\frac{-(5 / 2 c) \pm\left((5 / 2 c)^{2}+2 c^{2}\right)^{1 / 2}}{2}=\frac{\sqrt{33}-5}{4 c} \nonumber \]

Therefore the magnitudes of the two components are equal when

\[t_{2}=\sqrt{\frac{\sqrt{33}-5}{4 c}} \nonumber \]