Q 7.23    A man stands on a rotating platform, with his arms stretched horizontally holding a
              5kg  weight in each hand. The angular speed of the platform is 30 revolutions per
              minute. The man then brings his arms close to his body with the distance of each
              weight from the axis changing from 90 cm to 20 cm. The moment of inertia of the
              man together with the platform may be taken to be constant and equal to 7.6 kg m^{2}.

              (a) What is his new angular speed? (Neglect friction.)

Answers (1)

Moment of inertia when hands are stretched :

                                                                           =\ 2\times mr^2\ =\ 2\times (0.5)9^2

                                                                                                       =\ 8.1\ Kg\ m^2

So the moment of inertia of system (initial )  =   7.6  +  8.1  =  15.7 Kg m2.

Now, the moment of inertia when hands are folded :

                                                                                    =\ 2\times mr^2\ =\ 2\times 5(0.2)^2\ =\ 0.4\ Kg\ m^2

Thus net final moment of inertia is :     =   7.6  +  0.4  =  8 Kg m2.

Using conversation of angular momentum we can write :

                                                                      I_1\omega_1\ =\ I_2 \omega_2

or                                                                       \omega_2 \ =\ \frac{15.7\times 30}{8}\ =\ 58.88\ rev/min

                                                              

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