/*! 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 43 \(\|\) Ice skaters often end the... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\|\) Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well. Suppose that the moment of inertia of a skater with arms out and one leg extended is \(3.2 \mathrm{kg} \cdot \mathrm{m}^{2}\) and for arms and legs in is \(0.80 \mathrm{kg} \cdot \mathrm{m}^{2}\). If she starts out spinning at 5.0 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

Short Answer

Expert verified
The final angular speed of the skater, when her arms and one leg open outward, is 1.25 revolutions per second.

Step by step solution

01

Convert initial angular speed to rad/s

First, convert the initial angular speed of the skater from revolutions per second (rev/s) to radians per second (rad/s) by multiplying by \(2 \pi\), as one complete revolution equals \(2 \pi\) radians. So, \(w_{initial} = 5.0 rev/s \times 2 \pi rad/rev = 10 \pi rad/s.\)
02

Apply the conservation of angular momentum

Next, apply the conservation of angular momentum formula \((I_{initial} \cdot w_{initial} = I_{final} \cdot w_{final})\). Substituting the values: \(0.80 kg \cdot m^2 \times 10 \pi rad/s = 3.2 kg \cdot m^2 \times w_{final}\).
03

Solve for final angular speed

Rearranging the above equation to find the final angular velocity \(w_{final}\) gives: \(w_{final} = \frac{0.80 kg \cdot m^2 \times 10 \pi rad/s}{3.2 kg \cdot m^2} = \frac{5}{2} \pi rad/s.= 2.5 \pi rad/s.\)
04

Convert final angular speed to rev/s

Finally, convert the final angular speed to rev/s by dividing by \(2 \pi\), because \(2 \pi\) radians make one complete revolution. This gives: \(w_{final} = \frac{2.5 \pi rad/s}{2 \pi rev/rad} = 1.25 rev/s.\)

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Key Concepts

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

Angular Speed
Angular speed is a measure of how quickly an object is rotating. It is analogous to linear speed but instead of distance per unit time, angular speed refers to the angle through which a point or object turns in a certain amount of time. In rotational motion, angular speed is usually measured in revolutions per second (rev/s) or radians per second (rad/s). Angular speed is crucial in understanding rotational dynamics as it directly relates to how fast an object is spinning. The ice skater exercise beautifully illustrates the concept of angular speed by showing the change in rotation rate when the skater alters their body position. By extending their arms and leg, the skater decreases their angular speed due to conservation of angular momentum, which is a principle detailing that the angular momentum of a system remains constant if no external torques act on it.

Moment of Inertia
The moment of inertia, often symbolized by the letter 'I', is a measure of how much torque is required for a particular angular acceleration about a rotation axis. Think of it as the rotational equivalent of mass in Newton's second law of motion. The higher an object's moment of inertia, the harder it is to change its rotational speed. This property depends on both the mass of the body and the distribution of that mass in relation to the axis of rotation. Like mass in linear dynamics, moment of inertia plays a crucial role in how an object behaves in rotational motion. In the case of our spinning skater, with arms and a leg extended, the distribution of the skater's mass is farther from the center, which increases the moment of inertia. This, in turn, impacts how angular speed changes when the skater adjusts their body position.

In the exercise improvement advice, detailing the moment of inertia's role and its physical significance can aid in a student's intuitive understanding of the concept, underscoring not just the calculation, but also why and how moment of inertia affects rotational motion.
Angular Velocity
Angular velocity is a vector quantity that represents both the angular speed of an object and the axis around which the object is rotating. While angular speed is purely a scalar quantity, angular velocity also provides directional information. This distinction is useful in more complex physics problems, where the direction of rotation is important. In the context of our skater, the exercise focuses primarily on the magnitude of the angular velocity, which changes as the skater's configuration changes.

The conservation of angular momentum determines the skater's change in angular velocity. Since no external torques are mentioned in the problem, we can assume that the total angular momentum remains constant. Therefore, if the moment of inertia increases when the skater extends their limbs, the magnitude of angular velocity must decrease to maintain the same angular momentum, as shown in the exercise.
Rotational Motion
Rotational motion is the movement of an object in a circular path around a center or axis. Examples of rotational motion include a playground merry-go-round, the spinning of the Earth on its axis, and, as mentioned in our problem, a figure skater spinning. Key aspects of rotational motion involve torque, angular velocity, angular acceleration, and the moment of inertia. These elements are interconnected; torque causes angular acceleration, which changes angular velocity, and the moment of inertia defines how an object's mass is distributed as it rotates.

Through exercises such as the one with the ice skater, students get a clear demonstration of how these physical quantities interplay. Providing examples of rotational motion from everyday life, explaining how torque is applied to make things spin, and illustrating the conservation laws involved, can greatly improve a student's grasp of complex concepts like rotational dynamics.

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Most popular questions from this chapter

Two ice skaters, with masses of \(75 \mathrm{kg}\) and \(55 \mathrm{kg},\) stand facing each other on a \(15-\mathrm{m}\) -wide frozen river. The skaters push off against each other, glide backward straight toward the river's edges, and reach the edges at exactly the same time. How far did the 75 kg skater glide?

Casey is driving a \(1600 \mathrm{kg}\) car toward the east. She goes through an intersection at a speed of \(16 \mathrm{m} / \mathrm{s}\) (approximately \(35 \mathrm{mph}),\) the speed limit on both roads of the intersection. Kerry is driving a car of mass \(1200 \mathrm{kg}\) into the intersection, going north, and doesn't see or doesn't heed a red light, and slams into Casey's car. The cars lock together and skid to a stop. Later, the two review the scene with the police. Skid marks from the instant after the collision reveal that the two cars were moving exactly northeast. Kerry claims to have been driving at the speed limit, but Casey says that Kerry seemed to be going over the speed limit before the collision. Who is correct? Use the concept of conservation of momentum to make your case.

A small, 100 g cart is moving at \(1.20 \mathrm{m} / \mathrm{s}\) on a frictionless track when it collides with a larger, \(1.00 \mathrm{kg}\) cart at rest. After the collision, the small cart recoils at \(0.850 \mathrm{m} / \mathrm{s}\). What is the speed of the large cart after the collision?

\(\|\) At the county fair, Chris throws a \(0.15 \mathrm{kg}\) baseball at a \(2.0 \mathrm{kg}\) wooden milk bottle, hoping to knock it off its stand and win a prize. The ball bounces straight back at \(20 \%\) of its incoming speed, knocking the bottle straight forward. What is the bottle's speed, as a percentage of the ball's incoming speed?

\(A\) typical raindrop is much more massive than a mosquito and much faster than a mosquito flies. How does a mosquito survive the impact? Recent research has found that the collision of a falling raindrop with a mosquito is a perfectly inelastic collision. That is, the mosquito is "swept up" by the raindrop and ends up traveling along with the raindrop. Once the relative speed between the mosquito and the raindrop is zero, the mosquito is able to detach itself from the drop and fly away. a. A hovering mosquito is hit by a raindrop that is 40 times as massive and falling at \(8.2 \mathrm{m} / \mathrm{s},\) a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic? b. Because a raindrop is "soft" and deformable, the collision duration is a relatively long \(8.0 \mathrm{ms}\). What is the mosquito's average acceleration, in \(g\) 's, during the collision? The peak acceleration is roughly twice the value you found, but the mosquito's rigid exoskeleton allows it to survive accelerations of this magnitude. In contrast, humans cannot survive an acceleration of more than about \(10 \mathrm{g}\).

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