/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 The Spinning Figure Skater. The ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 (Fig. 10.49 ) 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 8.0 \(\mathrm{kg}\) . When outstretched, they span 1.8 \(\mathrm{m}\) ; when wrapped, they form a cylinder of radius 25 \(\mathrm{cm}\) . The moment of inertia about the rotation axis of the remainder of his body is constant and equal to 0.40 \(\mathrm{kg} \cdot \mathrm{m}^{2}\) . If his original angular speed is 0.40 \(\mathrm{rev} / \mathrm{s}\) , what is his final angular speed?

Short Answer

Expert verified
The final angular speed is approximately 1.14 rev/s.

Step by step solution

01

Compute initial moment of inertia of outstretched arms

When the arms are outstretched, they act like a slender rod pivoting about its center. The formula for the moment of inertia of a rod about its center is \( I = \frac{1}{12} M L^2 \). Substitute \( M = 8.0 \) kg and \( L = 1.8 \) m: \[ I_{rod} = \frac{1}{12} \times 8.0 \times (1.8)^2 = 2.16 \ \text{kg} \cdot \text{m}^2 \]
02

Compute initial total moment of inertia

Sum the moment of inertia of the rod with the constant inertia of the body: \[ I_{initial} = I_{rod} + I_{body} = 2.16 + 0.40 = 2.56 \ \text{kg} \cdot \text{m}^2 \]
03

Compute the moment of inertia of wrapped arms

When the arms are wrapped, they form a hollow cylinder. Use the formula \( I_{cylinder} = M R^2 \) for a hollow cylinder. Substitute \( M = 8.0 \) kg and \( R = 0.25 \) m: \[ I_{wrapped} = 8.0 \times (0.25)^2 = 0.50 \ \text{kg} \cdot \text{m}^2 \]
04

Compute final total moment of inertia

Sum the inertia of the hollow cylinder with the constant inertia of the body: \[ I_{final} = I_{wrapped} + I_{body} = 0.50 + 0.40 = 0.90 \ \text{kg} \cdot \text{m}^2 \]
05

Apply conservation of angular momentum

Angular momentum is conserved, so initial angular momentum equals final angular momentum. \[ L_{initial} = L_{final} \] which implies \[ I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final} \]
06

Solve for final angular speed

The initial angular speed is given as \( \omega_{initial} = 0.40 \ \text{rev/s} = 0.40 \times 2\pi \ \text{rad/s} \). Substitute into the conservation equation and solve for \( \omega_{final} \): \[ 2.56 \cdot (0.40 \times 2\pi) = 0.90 \cdot \omega_{final} \] Thus, \[ \omega_{final} = \frac{2.56 \cdot 0.40 \cdot 2\pi}{0.90} = 7.16 \ \text{rad/s} \] Convert back to revolutions per second: \[ \omega_{final} = \frac{7.16}{2\pi} \approx 1.14 \ \text{rev/s} \]

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Key Concepts

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

Moment of Inertia
Understanding the concept of moment of inertia is fundamental when dealing with rotational dynamics. Essentially, moment of inertia can be thought of as the rotational equivalent of mass in linear motion. It describes how much torque (rotational force) is needed for a desired angular acceleration about a rotational axis.

Just like mass tells us how much force we need to push an object in a straight line, the moment of inertia tells us how difficult it is to make something spin. It's calculated based on both the mass of an object and how the mass is distributed relative to the axis of rotation. This is why arms outstretched, like in our skater example, have a larger moment of inertia than when pulled in tightly.

For various shapes, this moment is calculated differently. Some key examples include:
  • For a slender rod, the formula is: \( I = \frac{1}{12} M L^2 \), where \( M \) is mass and \( L \) is length.
  • For a hollow cylinder, it's simply \( I = M R^2 \), reflecting its mass concentrated at a distance \( R \) from the axis.
These calculations allow us to better visualize and compute the mechanical behavior of rotating bodies.
Angular Speed
Angular speed refers to how quickly an object is rotating. It's often expressed in revolutions per second (rev/s) or radians per second (rad/s).

One complete revolution is equal to \( 2\pi \) radians, making conversions between these units straightforward. If a skater's initial angular speed is given in rev/s, converting to rad/s would involve multiplying by \( 2\pi \).

In the context of our skater, knowing both the initial and final angular speeds is crucial. With arms extended, less angular speed is needed to conserve angular momentum than when arms are pulled in very close to the body. As the moment of inertia decreases, angular speed must increase to conserve angular momentum, and vice versa. This conservation of angular momentum is what allows the skater to spin faster when pulling their arms in.
Rotational Dynamics
Rotational dynamics involves the study of forces and torques and their effect on motion in rotating systems.

In our spinning skater example, we utilize the conservation of angular momentum to understand the dynamics at play. Angular momentum \((L)\) is a key quantity in rotational dynamics, defined as the product of moment of inertia \((I)\) and angular speed \((\omega)\): \[ L = I \times \omega \]

This quantity is conserved as long as no external torques act on the system. So, initially, \( L_{initial} = I_{initial} \times \omega_{initial} \), and finally, \( L_{final} = I_{final} \times \omega_{final} \). Because these are equal, we can solve for any unknown quantity when managing a closed system.

This principle is why when a skater pulls their arms close, reducing their moment of inertia, their angular speed increases dramatically. This intuitive understanding of rotational dynamics allows for practical application in various fields, from astrophysics to mechanical engineering.

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Most popular questions from this chapter

A child rolls a 0.600 -kg basketball up a long ramp. The bas- ketball can be considered a thin-walled, hollow sphere. When the child releases the basketball at the bottom of the ramp, it has a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) . When the ball returns to her after rolling up the ramp and then rolling back down, it has a speed of 4.0 \(\mathrm{m} / \mathrm{s}\). Assume the work done by friction on the basketball is the same when the ball moves up or down the ramp and that the basketball rolls without slipping. Find the maximum vertical height increase of the ball as it rolls up the ramp.

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