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I The spinning figure skater. The outstretched hands and arms of a figure skater preparing for a spin can be considered a slender rod pivoting about an axis through its center. (See Figure \(10.57 .)\) When the skater's hands and turms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered a thinwalled hollow cylinder. His hands and arms have a combined mass of \(8.0 \mathrm{~kg}\). When outstretched. they span \(1.8 \mathrm{~m}\); when wrapped, they form a cylinder of radius 25 cm. The moment of inertia about the axis of rotation of the remainder of his body is constant and equal to \(0.40 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If the skater's original angular speed is \(0.40 \mathrm{rev} / \mathrm{s},\) what is his final angular speed?

Step by step solution

01

Calculate Initial Moment of Inertia

The initial moment of inertia when the arms are outstretched is calculated as a rod pivoting about the center. The formula used is \( I = \frac{1}{12} m L^2 \), where \( m \) is the mass of the arms and \( L \) is the span when outstretched. Substituting the values: \( I_1 = \frac{1}{12} \times 8.0 \times (1.8)^2 = 2.16 \ \mathrm{kg \cdot m^2} \). Adding the moment of inertia of the body: \( I_{\text{total, initial}} = I_1 + 0.40 = 2.56 \ \mathrm{kg \cdot m^2} \).

02

Calculate Final Moment of Inertia

When the arms are wrapped, they form a hollow cylinder. The moment of inertia for a thin-walled hollow cylinder is \( I = m r^2 \), where \( r \) is the radius of the cylinder. Using this formula: \( I_2 = 8.0 \times (0.25)^2 = 0.50 \ \mathrm{kg \cdot m^2} \). Adding the moment of inertia of the body: \( I_{\text{total, final}} = I_2 + 0.40 = 0.90 \ \mathrm{kg \cdot m^2} \).

03

Apply Conservation of Angular Momentum

According to the conservation of angular momentum, the initial and final angular momentum must be equal: \( I_{\text{total,initial}} \times \omega_i = I_{\text{total,final}} \times \omega_f \). Given the initial angular speed \( \omega_i = 0.40 \ \mathrm{rev/s} \) (convert to radians per second: \( 0.40 \times 2\pi \)), solve for \( \omega_f \): \[2.56 \times 0.40 \times 2\pi = 0.90 \times \omega_f\]which simplifies to \( \omega_f \approx \frac{2.56 \times 0.40 \times 2\pi}{0.90} \).

04

Calculate Final Angular Speed

Finding \( \omega_f \):\[\omega_f \approx \frac{2.56 \times 0.40 \times 2\pi}{0.90} \approx 7.15 \ \mathrm{rad/s}\]Convert the final speed back into revolutions per second:\[\omega_f = \frac{7.15}{2\pi} \approx 1.14 \ \mathrm{rev/s}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia

The moment of inertia is crucial in understanding rotational dynamics. It tells us how difficult it is for an object to change its rotational state. This concept is similar to mass in linear motion. For a rod, when arms are outstretched, the formula used is \( I = \frac{1}{12} mL^2 \), where:

• \( m \) is the mass of the rod.
• \( L \) is the length of the rod.

When the mass is closer to the axis of rotation, it becomes easier for the skater to spin faster due to a smaller moment of inertia. This physical disposition alters when arms are tucked in, transforming the system into a thin-walled cylinder.

Rotational Dynamics

Rotational dynamics explains how forces affect rotational motion. It acts as the equivalent to Newton's Second Law in terms of rotation. Here, the force applied by the skater's muscles translates into torque, which is the rotational equivalent of force. The relationship between torque (\( \tau \)), moment of inertia (\( I \)), and angular acceleration (\( \alpha \)) is given by \( \tau = I \alpha \).

In this problem, as the arms pull inwards, the torque remains constant, but the change in moment of inertia allows for greater angular velocity, showcasing how rotational dynamics govern stability and speed in a spinning body.

Angular Velocity

Angular velocity refers to how fast something rotates or spins. It is denoted by \( \omega \) and is measured in radians per second or revolutions per second. Initially, the skater's angular velocity was \( 0.40 \) revolutions per second.

When arms are pulled in, the conservation of angular momentum principle ensures that any decrease in moment of inertia results in an increase in angular velocity. Thus, the final angular velocity becomes \( 1.14 \) revolutions per second. This change is crucial for performers seeking high-speed rotations.

Thin-Walled Cylinder

When considered as a thin-walled hollow cylinder, the skater's arms wrapped around the body have a new moment of inertia representation: \( I = mr^2 \), where \( r \) is the radius.

In this scenario:

• The radius \( r \) is \( 0.25 \) meters, representing the distance from the pivot to the edges of the hollow cylinder formed by the arms.
• The mass \( m \) remains constant at \( 8.0 \) kg.

The thin-walled cylinder formula reflects the decreased resistance to rotational change, indicating how compacting mass affects rotational characteristics.

Physics Problem Solving

Physics problem solving requires understanding and applying multiple concepts like moment of inertia and conservation of momentum. Approach such problems with clear steps:

• Identify all known variables.
• Use appropriate formulas for the moment of inertia based on object shapes.
• Apply the conservation laws to relate initial and final states.

Ensuring that units are consistent, like converting revolutions per second to radians, is vital for accuracy. Mastery of these systematic approaches makes it easier to untangle complex rotational dynamics problems.