Question

On average, both arms and hands together account for 13 % of a person's mass, while the head is 7.0% and the trunk and legs account for 80 % . We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 61.0-kg skater is 1.80 m tall, has arms that are each 70.0 cm long (including the hands), and a trunk that can be modeled as being 35.0 cm in diameter. The skater is initially spinning at 70.0 rpm with his arms outstretched.
Required:
What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?

Answer 1

Answer:

$\dot n_{f} = 85.177\,rpm$

Explanation:

The expression for the moment of inertia of the person is:

Arms outstretched

$I = \frac{1}{12}\cdot (0.13)\cdot (61\,kg)\cdot (1.40\,m)^{2} + \frac{1}{2}\cdot (0.87)\cdot (61\,kg)\cdot (0.35\,m)^{2}$

$I = 4.546\,kg\cdot m^{2}$

Arms parallel to the trunk

$I = \frac{1}{2}\cdot (61\,kg)\cdot (0.35\,m)^{2}$

$I = 3.736\,kg\cdot m^{2}$

The final angular speed is found by means of the Principle of Angular Momentum Conservation:

$I_{o}\cdot \dot n_{o} = I_{f}\cdot \dot n_{f}$

$\dot n_{f} = \frac{I_{o}}{I_{f}}\cdot \dot n_{o}$

$\dot n_{f} = \left(\frac{4.546\,kg\cdot m^{2}}{3.736\,kg\cdot m^{2}}\right)\cdot (70\,rpm)$

$\dot n_{f} = 85.177\,rpm$