Question

A uniform circular disk of moment of inertia 8.0 kg.m² is rotating at 4.0 rad/s. A small lump of mass 1.0 kg is dropped on the disk and sticks to it at a distance of 1.0 m from the axis of rotation.
What is the new rotational speed of the combined system?

Answer 1

Answer:

$\omega'=32\ rad.s^{-1}$

Explanation:

Given:

• moment of inertial of a uniform circular disk, $I=8\ kg.m^2$

• angular speed of rotation, $\omega=4\ rad.s^{-1}$

• Mass of lump, $m'=1\ kg$

• position of the lump from the center of the disk, $r'=1\ m$

Using the conservation of the angular momentum:

$I.\omega=I'.\omega'$

$8\times 4=m'.r'^2\times \omega'$

$32=1\times 1\times\omega'$

$\omega'=32\ rad.s^{-1}$