Question

The sphere of mass M and radius R in FIGURE P12.76 is rigidly attached to a thin rod of radius r that passes through the sphere at distance 12 R from the center. A string wrapped around the rod pulls with tension T. Find an expression for the sphere’s angular acceleration. The rod’s moment of inertia is negligible.

Short answer

The expression for the sphere’s angular acceleration is $\alpha =\frac{20Tr}{13M{R}^2}$

Step-by-step solution

Given information

Mass of sphere = M

Radius of sphere = R
radius of rod = r
Sphere is attached with Rod at R/2 distance from the center of the sphere.
Rod is pulled with a string with tension T
The moment of inertia of the rod is negligible.

Explanation

From the parallel axis theorem moment of inertia is given by

$$\begin{gathered} I=\frac{2}{5}M{R}^2+M{\left(\frac{R}{2}\right)}^2 \\ I=M{R}^2\left(\frac{2}{5}+\frac{1}{4}\right) \\ I=\frac{13M{R}^2}{20} \end{gathered}$$

Torque on the sphere is calculated by

$\tau =Fr\sin \theta$

In this case only tension force is applicable so

$\tau =Tr............................\left(2\right)$

Torque is calculated by another way using angular acceleration

$\tau =I\alpha ..........................\left(3\right)$

From equation (2) and (3)

$$\begin{gathered} I\alpha =Tr \\ \alpha =\frac{Tr}{I}.......................\left(4\right) \end{gathered}$$

Substitute values , we get

$$\begin{gathered} \alpha =\frac{Tr}{\left(\frac{13M{R}^2}{20}\right)} \\ \alpha =\frac{20Tr}{13M{R}^2} \end{gathered}$$