Question

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings her arms inward toward her body in such a way that the distance of each mass from the axis changes from 1.00 to 0.50 m. Her rate of rotation (neglecting the mass of the skater) will the final angular velocity (in rev/s) be if her moment of inertia remains approximately constant at 5kg(m2), and the distance of the masses from the axis changes from 1 m to 0.1m?a. 6b. 3c. 9d. 4e. 7

Answer 1

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    $\tau = I \times \alpha$

and,       I = $\sum mr^{2}$

So,      $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

                       = 4 $kg m^{2}$

      $\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

     $\tau_{2} = I_{2} \alpha_{2}$

So,      $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

                     = 1 $kg m^{2}$

 as $\tau_{2} = I_{2} \alpha_{2}$

                   = $1 kg m^{2} \times \alpha_{2}$        

Hence,     $\tau_{1} = \tau_{2}$

                  $4 \alpha_{1} = \alpha_{2}$

            $\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.