Answer to Question #128348 in Classical Mechanics for Hassam

Question

A thin, very light wire is wrapped
around a drum that is free to rotate.
The free end of the wire is attached
to a ball of mass m. The drum has
the same mass m. Its radius is R and
its moment of inertia is I = (1/2)mR2.
As the ball falls, the drum spins.
At an instant that the ball has
translational kinetic energy K, then what rotational kinetic energy does the drum has? explain.

Answer 1

Rotational kinetic energy is expressed as $k(rotational)=\frac{1}{2}lw^2$

$Kdrum=\frac{1}{2}lw^2(\frac{1}{2}mR^2)(\frac{v}{R})^2=\frac{1}{4}mv^2=\frac{1}{2}K$

The energy which is due to rotation of an object and makes part of its total kinetic energy refers to as rotational kinetic energy.