Question

The Spinning Figure Figure E10.43 Skater. The outstretched hands and arms of a figure skater preparing for a spin can be con- sidered a slender rod pivoting about an axis through its center (Fig. E1043). When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thin- walled, hollow cylinder. His hands and arms have a combined mass of 8.0 kg. When outstretched, they span 1.8 m; when wrapped, they form a cylinder of radius 25 cm. The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 kg.m. If his original angular speed is 0.40 rev/s, what is his final angular speed?

Answer 1

To calculate the final angular speed of the skater, we can use the conservation of angular momentum. By calculating the initial and final moments of inertia, we can find the final angular speed.

Step-by-step explanation:

To calculate the final angular speed of the skater, we can use the conservation of angular momentum. Initially, the skater has an initial angular speed of 0.40 rev/s. When his hands and arms are brought in and wrapped around his body, the moment of inertia of his hands and arms decreases. The moment of inertia is directly proportional to the square of the radius. So, by calculating the initial and final moments of inertia, we can find the final angular speed using the conservation of angular momentum equation:

{{insert formula here}}

Answer 2

His final angular speed is
`0.78 \, \text{rev/s}\)`

How to determine his final speed?

Given:

`\(L_w = 0.4\), \(I_w = 0.8032 \, \text{kgm}^2\)`

For the stretched arms:

`\(I_s = 0.4 + I_p\)`

`\(I_p = (ML^2)/(12)\) (Given \(M = 7 \, \text{kg}\), \(L = 1.7 \, \text{m}\))`

`\(I_p = (7 * 1.7 * 1.7)/(12) = 1.6858 \, \text{kgm}^2\)`

`\(I_s = 0.4 + 1.6858 = 2.0858 \, \text{kgm}^2\)`

Using conservation of angular momentum:

Initial angular momentum
`= \(I_s * \omega_i = 2.0858 * 0.30 = 0.62574 \, \text{Kg-m}^2\text{-rev/s}\)`

Let the final angular velocity be
`\(\omega_f\)`.

Using conservation of angular momentum:
`\(0.62574 = I_w * \omega_f\)`

`\(0.8032 * \omega_f = 0.62574\)`

Therefore:
`\(\omega_f = (0.62574)/(0.8032) = 0.7790 \, \text{rev/s}\)` or rounded to
`\(\omega_f = 0.78 \, \text{rev/s}\)`.