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  • 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.

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.

Answers (1)

best_answer

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

Posted by

Devendra Khairwa

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