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# CALC Mass M is distributed uniformly over a disk of radius

ISBN: 9780321675460 31

## Solution for problem 89CP Chapter 13

University Physics | 13th Edition

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Problem 89CP

CALC Mass M is distributed uniformly over a disk of radius a . Find the gravitational force (magnitude and direction) between this disk-shaped mass and a particle with mass m located a distance x above the center of the disk (Fig. P13.81). Does your result reduce to the correct expression as x becomes very large? (Hint: Divide the disk into infinitesimally thin concentric rings, use the expression derived in Exercise 13.35 for the gravitational force due to each ring, and integrate to find the total force.) 13.35 CALC Consider the ring- shaped body of Fig. E13.35. A particle with mass m is placed a distance x from the center of the ring, along the line through the center of the ring and perpendicular to its plane. (a) Calculate the gravitational potential energy U of this system. Take the potential energy to be zero when the two objects are far apart. (b) Show that your answer to part (a) reduces to the expected result when x is much larger than the radius a of the ring. (c) Use Fx = -dU/dx to find the magnitude and direction of the force on the particle (see Section 7.4). (d) Show that your answer to part (c) reduces to the expected result when x is much larger than a. (e) What are the values of U and Fx when x = 0? Explain why these results make sense.

Step-by-Step Solution:

Solution 89CP Introduction We have to find out the expression for the gravitational force between the point mass and the disc. Then we have to use the approximation x a and show that the expression leads to the correct expression for gravitational force between two particle. Step 1 The gravitational force between the ring of radius r and mass m and the distance from the ring r to the point mass x is given by F = Gmm r ……………………………………..(1) r (x +r )2 Step 2 Let us now consider a thin circular ring on the disc with radius r and thickness dr. If is the surface mass density of the disc, then the mass of the ring is m r dM = 2rdr We know that M is the total mass of the disc and a is the radius of the disc, hence the surface mass density is given by = a2 Hence using the above equation we can write that M 2M dM = r22rdr = a2rdr

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