A figure skater weighing 70 kg begins to rotate with a frequency of 1 rpm

A figure skater weighing 70 kg begins to rotate at a frequency of 1 rps, holding his arms horizontally. At what frequency will he rotate if he raises his arms vertically? The figure skater’s body is considered to be a homogeneous cylinder with a radius of 15 cm, the arms are considered to be rods of 0.75 m and a mass of 5 kg each.

Let's imagine the skater as a homogeneous cylinder of radius $r = 15,\text{cm}$ and mass $m_1 = 70,\text{kg}$. Then the moment of inertia of his body relative to the vertical axis passing through his center of mass is equal to:

$$I_1 = \frac{1}{2}m_1r^2 = 0.07875 ,\text{кг}\cdot\text{м}^2$$

When a skater holds his arms horizontally, his moment of inertia increases due to the moments of inertia of the arms. The moment of inertia of each arm, considering it a thin rod, about a vertical axis passing through its end is equal to:

$$I_2 = \frac{1}{3}m_2l^2 = 0.703125 ,\text{кг}\cdot\text{м}^2$$

where $m_2 = 5 ,\text{kg}$ is the mass of each arm, and $l = 0.75 ,\text{m}$ is the length of each arm.

Thus, the moment of inertia of the system of the skater and his arms, when he holds them horizontally, is equal to:

$$I_0 = I_1 + 2I_2 = 1.484 ,\text{кг}\cdot\text{м}^2$$

When a skater raises his arms vertically, his moment of inertia decreases. The moment of inertia of each arm about a vertical axis passing through its center of mass is:

$$I_3 = \frac{1}{12}m_2l^2 + \frac{1}{4}m_2r^2 = 0.134766 ,\text{кг}\cdot\text{м}^2$$

Thus, the moment of inertia of the skater’s system and his arms when he raises them vertically is equal to:

$$I_4 = I_1 + 2I_3 = 0.34828 ,\text{кг}\cdot\text{м}^2$$

According to the law of conservation of angular momentum $I\omega = const$, where $\omega$ is the angular velocity of rotation, we obtain:

$$I_0\omega_0 = I_4\omega_4$$

From here you can find the desired angular velocity $\omega_4$ at which the skater will rotate while holding his arms vertically:

$$\omega_4 = \frac{I_0}{I_4}\omega_0 = \frac{1.484}{0.34828}\cdot 1 = 4.26 ,\text{об/с}$$

Thus, when a skater raises his arms vertically, his rotational speed will increase by 4.26 times.

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This item is problem 11088, which describes the motion of a 70 kg figure skater starting a 1 rps spin with his arms horizontal, and requires him to determine at what speed he would spin if he raised his arms vertically.

To solve the problem it is necessary to use the laws of conservation of energy and angular momentum. The figure skater's body is considered a homogeneous cylinder with a radius of 15 cm, and the arms are considered to be rods of 0.75 m and a mass of 5 kg each.

First, it is necessary to determine the moment of inertia of the skater’s system and his arms relative to the axis of rotation. Then, using the law of conservation of angular momentum, we can find the angular velocity of the skater's rotation after raising his arms. Using the law of conservation of energy, you can find the speed of the skater's rotation.

The solution to the problem is described in detail in the text of task 11088, which contains the formulas and laws used for the solution, the derivation of the calculation formula and the answer. If you have any questions about the solution, you can ask them and get help.

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