Question

4. A boy stands at the centre of a turn table with his two arms stretched. The turn table is set rotating with an angular with 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 2/5times the initial? 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 correct answer is 100 rpm.

Explanation:

Use the conservation of angular momentum.

Iw= I'w'

Here, I is the initial moment of the inertia of the object, w is the initial angular velocity, I' is the final moment of inertia of the object and w' is the final angular velocity.

It is given in the problem that the turn table is set rotating with an angular with an angular speed of 40 r.p.m. The final moment of inertia reduces to 2/5 times the initial.

Put w= 40 rpm and $I'=\frac{2}{5}I$.

$I(40)= w'(\frac{2}{5}I)$

$w'=100 rpm$

Therefore, the final moment of inertia of the boy is 100 rpm.