Answer to Question #281893 in Physics for Fahad

Question

A solid sphere of mass 2 kg and a massless string of total length 4 meters which is wrapped around the equator of the sphere (total length includes the length of wrapped string and the length of the string hanging from the ceiling). The sphere then falls vertically while rolling down (without slipping) the whole wrapped string and falls on top of an inclined plane. Due to the collision with the inclined plane, it loses all its kinetic energy within a very short time and starts rolling along the inclined plane with zero initial velocity. Consider that a total of 10 Joules of energy is lost during the motion of the sphere along the planes.The magnitude of gravitational acceleration g = 9.8 𝑚/𝑠2. (a) Find the value of the moment of inertia of the sphere about its axis of rotation.(b) What is the magnitude of angular velocity of the sphere when it touches the inclined plane?(c) How far along the second inclined plane (45º) on the right will the center of the sphere travel before it comes to a stop?

Answer 1

(a) Find the value of the moment of inertia of the sphere about its axis of rotation:

$$I=\frac25mr^2=0.8r^2,$$

where r is the radius of the sphere.

(b) The magnitude of angular velocity of the sphere when it touches the inclined plane:

$$mgy=\frac12mv^2+\frac12·\frac25mr^2·\bigg(\frac vr\bigg)^2,\\\space\\ v=\sqrt{10gy/7.}$$

(c) The length can be found from the following relation:

$$mgh\sin30°-10=mgL\sin45°,\\\space\\ L=4.9\text{ m}.$$