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Textbook Solutions for Sears and Zemansky's University Physics with Modern Physics

Chapter 14 Problem 94P

Question

Problem 94P

CP Two uniform solid spheres, each with mass M = 0.800 kg and radius R = 0.0800 m, are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant k = 160 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

Solution

Solution 94P

Introduction

We have to show that the motion of the system is simple harmonic. Then we have to calculate the period of motion.

We will first calculate the total energy of the system and then using the conservation of energy we can show that the motion is simple harmonic.

Step 1

The kinetic energy of the each ball is given by

……………………(1)

Where  is the moment of inertia and  is the angular speed of the rotation of the ball.

Now the moment of inertia of sphere is given by

Where  is the radius of the sphere.

The angular speed is given by

Putting the above values in equation (1) we have

     

     

So the total kinetic energy of the two ball is given by

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full solution

Title Sears and Zemansky's University Physics with Modern Physics 13 
Author Hugh D. Young; Roger A. Freedman; A. Lewis Ford
ISBN 9780321696861

Solved: CP Two uniform solid spheres, each with mass M =

Chapter 14 textbook questions

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