# Q7.13 (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.

We are given with the initial angular speed and the relation between the moment of inertia of both the cases.

Here we can use conservation of angular momentum as no external force is acting the system.

So we can write :

$I_1w_1\ =\ I_2w_2$

$w_2\ =\ \frac{I_1w_1}{I_2}$

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

$=\ 100\ rev/min$

## Related Chapters

### Preparation Products

##### Knockout NEET July 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
##### Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
##### Knockout JEE Main July 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Test Series NEET July 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-