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Chapter 10: Problem 43

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. Elo.43). 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 \(\mathrm{kg} .\) 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 of the skater is approximately 2.84 rad/s.

Step by step solution

01

Analyze the Problem

We need to find the final angular speed of the skater after his arms are wrapped around his body. Initially, the skater's arms are outstretched, and then they are wrapped around, changing the distribution of mass and thus the moment of inertia.
02

Understand Moment of Inertia

For the arms stretched out as a slender rod pivoting about its center: \[ I_{arms, stretched} = \frac{1}{12} m L^2 \] where:- \( m = 8.0 \, \mathrm{kg} \) (mass of arms),- \( L = 1.8 \, \mathrm{m} \) (length when outstretched).
03

Calculate Initial Moment of Inertia

Using the formula for a slender rod, compute the moment of inertia of the arms when outstretched:\[ I_{arms, stretched} = \frac{1}{12} \times 8.0 \, \mathrm{kg} \times (1.8 \, \mathrm{m})^2 \] \[ I_{arms, stretched} = 2.16 \, \mathrm{kg} \cdot \mathrm{m}^2 \] The total initial moment of inertia is:\[ I_{initial} = I_{arms, stretched} + I_{body} = 2.16 \, \mathrm{kg} \cdot \mathrm{m}^2 + 0.40 \, \mathrm{kg} \cdot \mathrm{m}^2 \] \[ I_{initial} = 2.56 \, \mathrm{kg} \cdot \mathrm{m}^2 \]
04

Understand Changed Moment of Inertia

After wrapping the arms, they form a hollow cylinder. Its moment of inertia is:\[ I_{arms, wrapped} = m r^2 \] where:- \( m = 8.0 \, \mathrm{kg} \) (mass of arms),- \( r = 0.25 \, \mathrm{m} \) (radius of cylinder).
05

Calculate Final Moment of Inertia

Using the formula for a hollow cylinder, compute the wrapped moment of inertia:\[ I_{arms, wrapped} = 8.0 \, \mathrm{kg} \times (0.25 \, \mathrm{m})^2 \] \[ I_{arms, wrapped} = 0.5 \, \mathrm{kg} \cdot \mathrm{m}^2 \] The total final moment of inertia is:\[ I_{final} = I_{arms, wrapped} + I_{body} = 0.5 \, \mathrm{kg} \cdot \mathrm{m}^2 + 0.40 \, \mathrm{kg} \cdot \mathrm{m}^2 \] \[ I_{final} = 0.9 \, \mathrm{kg} \cdot \mathrm{m}^2 \]
06

Apply Conservation of Angular Momentum

Angular momentum before and after the arms are wrapped remains constant:\[ I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final} \] where:- \( \omega_{initial} = 0.40 \, \mathrm{rev/s} \times \frac{2\pi}{1} \times \frac{1}{1} \, \mathrm{rad/s} = 0.40 \times 2\pi \, \mathrm{rad/s} \)- Convert revolutions per second to radians per second for calculations.
07

Solve for Final Angular Speed

Substitute known values into the conservation equation:\[ 2.56 \cdot 0.40 \times 2\pi = 0.9 \cdot \omega_{final} \] Solve for \( \omega_{final} \):\[ \omega_{final} = \frac{2.56 \cdot 0.40 \times 2\pi}{0.9} \] Calculate \( \omega_{final} \):\[ \omega_{final} \approx 2.84 \, \mathrm{rad/s} \]

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Key Concepts

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

Moment of Inertia
The concept of Moment of Inertia is crucial in understanding rotational motion. It refers to how much resistance an object has to changes in its rotation. Think of it like mass in linear motion, but for spinning or rotational movement. It depends not only on the mass of the object but also on how that mass is distributed relative to the axis of rotation.In the case of the figure skater, when his arms are outstretched, they can be modeled as a slender rod. The formula for the moment of inertia of a slender rod rotating about its center is \( I = \frac{1}{12} mL^2 \), where \( m \) is the mass and \( L \) is the length of the rod. When he wraps his arms, they turn into a hollow cylinder, for which the moment of inertia becomes \( I = mr^2 \), where \( r \) is the radius. This change in moment of inertia is what affects the skater's angular speed eventually.
Rotational Dynamics
Rotational Dynamics involves the analysis of rotational motion. It's akin to the linear dynamics we learn in basic physics, extending it to rotating objects. The key principle here is the conservation of angular momentum. Like linear momentum, angular momentum must be conserved unless acted upon by an external force. For the figure skater, his total angular momentum remains constant before and after he tucks his arms. This is represented mathematically as:
  • Initial angular momentum: \( L_i = I_{initial} \cdot \omega_{initial} \)
  • Final angular momentum: \( L_f = I_{final} \cdot \omega_{final} \)
Because no external torques are acting on him, \( L_i \) equals \( L_f \). This relationship helps calculate the skater鈥檚 new angular speed after altering his moment of inertia by changing the position of his arms.
Angular Speed Calculation
Calculating angular speed involves recognizing how changes in moment of inertia affect rotational speed due to conservation laws.For the figure skater, his initial angular speed is given (\(0.40 \text{ rev/s}\)), which needs to be converted into the standard unit, radians per second, for calculations. Using the conversion \(1 \text{ rev} = 2\pi \text{ rad}\), the initial speed in radians per second is \(0.40 \times 2\pi \). To find his final angular speed, \( \omega_{final}\), we use the equation from the conservation of angular momentum: \[ I_{initial} \cdot \omega_{initial} = I_{final} \cdot \omega_{final} \]Substituting the respective values, you solve for \( \omega_{final} \), illustrating how the skater's rotational speed increases as his moment of inertia decreases when he pulls his arms inwards.

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Most popular questions from this chapter

An experimental bicycle wheel is placed on a test stand so that it is free to turn on its axle. If a constant net torque of 7.00 \(\mathrm{N} \cdot \mathrm{m}\) is applied to the tire for 2.00 \(\mathrm{s}\) , the angular speed of the tire increases from 0 to 100 rev/min. The external torque is then removed, and the wheel is brought to rest by friction in its bearings in 125 s. Compute (a) the moment of inertia of the wheel about the rotation axis; (b) the friction torque; (c) the total number of revolutions made by the wheel in the 125 -s time interval.

A stone is suspended from the free end of a wire that is wrapped around the outer rim of a pulley, similar to what is shown in Fig. \(10.10 .\) The pulley is a uniform disk with mass 10.0 \(\mathrm{kg}\) and radius 50.0 \(\mathrm{cm}\) and turns on frictionless bearings. You measure that the stone travels 12.6 \(\mathrm{m}\) in the first 3.00 s starting from rest. Find (a) the mass of the stone and (b) the tension in the wire.

A small block with mass 0.250 \(\mathrm{kg}\) is attached to a string passing through a hole in a frictionless, horizontal surface (see Fig. E10.42). The block is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 \(\mathrm{m} / \mathrm{s}\) . The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 \(\mathrm{N} .\) What is the radius of the circle when the string breaks?

Rolling Stones. A solid, uniform, spherical boulder starts from rest and rolls down a \(50.0-\mathrm{m}\) -high hill, as shown in Fig. Plo.81. The top half of the hill is rough enough to cause the boulder to roll without slipping, but the lower half is covered with ice and there is no friction. What is the translational speed of the boulder when it reaches the bottom of the hill?

A Gyroscope on the Moon. A certain gyroscope precesses at a rate of 0.50 \(\mathrm{rad} / \mathrm{s}\) when used on earth. If it were taken to a lunar base, where the acceleration due to gravity is \(0.165 g,\) what would be its precession rate?
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