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

`\alpha= (2rF)/(3mR)`

Step-by-step 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 ( (R)/(2) \right )^2+mR^(2)`

=
`(3)/(2) mR^(2)`

now Substitute I=
`(3)/(2) mR^(2)` in the (1) we get

`rF= (3)/(2)mR^(2)\alpha`

rearranging for alpha we get

`\alpha= (2rF)/(3mR)`

hence the acceleration is

`\alpha= (2rF)/(3mR)`