Question

(a) A child stands at the centre of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of 40 rev/min. How much is the angular speed of the child if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the initial value ? Assume that the turntable rotates without friction.
(b) Show that the childs new kinetic energy of rotation is more than the initial kinetic energy of rotation. How do you account for this increase in kinetic energy?

Answer 1

(a) Since no external force acts on the boy, the angular momentum is a constant.

Thus $I_1{\omega }_1=I_2{\omega }_2$

$⟹{\omega }_2=\frac{I_1}{I_2}{\omega }_1=\frac{5}{2}\times 40rev/min=100revmin$

(b) Initial kinetic energy=$E_1=\frac{1}{2}{I}_1{\omega }_1^2$

Final kinetic energy=$E_2=\frac{1}{2}{I}_2{\omega }_2^2$

Thus $\frac{E_2}{E_1}=\frac{I_2}{I_1}\times \frac{{\omega }_2^2}{{\omega }_1^2}=2.5$

The increase in the rotational kinetic energy is attributed to the internal energy of the boy.