Question

If the sphere on the left is moved closer to the central cylinder and placed at a distance R/2 from the axis of rotation, what is the magnitude of the angular acceleration α of the modified system? Assume that the rest of the system doesn't change.

Answer 1

Answer:

$\alpha= \frac{2rF}{3mR}$

Explanation:

The torque reqired to rotate the above masses rod system

$\tau = rF= I\alpha$...............(1)

Total moment of inertia of masses and rod is

$I=2m\left ( \frac{R}{2} \right )^2+mR^{2}$

=$\frac{3}{2} mR^{2}$

now Substitute I=$\frac{3}{2} mR^{2}$ in the (1) we get

$rF= \frac{3}{2}mR^{2}\alpha$

rearranging for alpha we get

$\alpha= \frac{2rF}{3mR}$

hence the acceleration is

$\alpha= \frac{2rF}{3mR}$