Question

A boy stands at the centre of a turn table with his two arms stretched. The table is set rotating an angular speed of 40 r.p.m. How much is the angular speed of the boy if he folds his hands back and thereby reduces his moment of inertia to 2/5 times the inertial? Assume the turn table to rotate without friction.

Answer 1

moment of inertia * angular speed is always constant.w=40rpm is given and w' is to be found.
So,Iw=(Iw)'
or,I*40=(2/5)I*w'
hence,w=100rpm .

Answer 2

Answer:

The angular speed of the boy if he folds his hand is 100 r.p.m

Explanation:

Given that,

Initial angular speed = 40 r.p.m

The moment of inertia of the boy with stretched hand = I₁

The moment of inertia of the boy with folded hand = I₂

The two moment inertia are released as:

$I_{2}=\dfrac{2}{5}I_{1}$

No external force on the boy so the angular momentum is constant.

According to conservation of angular momentum

$L\ I_{1}\omega_{1}=L\ I_{2}\omega_{2}$

$I_{1}\omega_{1}=I_{2}\omega_{2}$

$\omega_{2}=\dfrac{I_{1}}{I_{2}}\omega_{1}$

$\omega_{2}=\dfrac{I_{1}}{\dfrac{2}{5}I_{1}}\times40$

$\omega_{2}=\dfrac{5}{2}\times40$

$\omega_{2}=100 r.p.m$

Hence, The angular speed of the boy if he folds his hand is 100 r.p.m