/*! 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 44 Force on a Skater's Wrist. A \(5... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Force on a Skater's Wrist. A \(52 \mathrm{~kg}\) ice skater spins about a vertical axis through her body with her arms horizontally outstretched; she makes 2.0 turns each second. The distance from one hand to the other is \(1.50 \mathrm{~m} .\) Biometric measurements indicate that each hand typically makes up about \(1.25 \%\) of body weight. (a) Draw a free-body diagram of one of the skater's hands. (b) What horizontal force must her wrist exert on her hand? (c) Express the force in part (b) as a multiple of the weight of her hand.

Short Answer

Expert verified
The horizontal force the wrist must exert on her hand is computed in Step 3. The force expressed as a multiple of the weight of her hand is computed in Step 4.

Step by step solution

01

Analyze the rotational motion

Firstly, we will compute the angular velocity of the skater. Since she makes 2.0 turns each second, the angular velocity is \(2.0 \, \text{revs/s} = 2.0 \times 2\pi \, \text{rad/s} = 4\pi \, \text{rad/s}.\) The radius of the circular motion is \(\frac{1.50 \, \mathrm{m}}{2} = 0.75 \, \mathrm{m}\).
02

Calculate the mass of the skater's hand

Each hand typically constitutes 1.25% of the skater's body weight. So, the mass of each hand is \((1.25 / 100) \times 52 \, \mathrm{kg} = 0.65 \, \mathrm{kg}.\)
03

Compute the horizontal force

By considering the forces acting on the skater's hand, we can see that the centripetal force causing the circular motion is provided by the horizontal force exerted by the skater's wrist. We can use the formula for centripetal force: \(F = m \omega^2 r\). Substituting the values we have, \(F = 0.65 \, \mathrm{kg} \times (4\pi \, \mathrm{rad/s})^2 \times 0.75 \, \mathrm{m}.\)
04

Express the force as a multiple of hand weight

The weight of the skater's hand is \(m \times g = 0.65 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s}^2\). Divide the computed force in Step 3 by the weight of the hand to get the force as a multiple of the hand weight.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Angular Velocity in Rotational Motion
Angular velocity is a pivotal concept in the study of rotational motion, representing how quickly an object rotates or revolves around a central axis. In the context of a spinning ice skater, as in our exercise, angular velocity quantifies the number of turns made per second. This is often measured in revolutions per second (rev/s), but for calculations, it's converted into radians per second (rad/s) because radians are the standard unit of angular measure in physics.

For example, our skater completes 2.0 turns each second, equating to an angular velocity of \(4\text{Ï€} \text{rad/s}\). This conversion is crucial for calculating other aspects of rotational motion, such as centripetal force. It's important to note that a complete revolution corresponds to \(2\text{Ï€} \text{rad}\), explaining the step in the calculation converting revolutions to radians.
Free-Body Diagrams: Visualizing Forces in Physics
A free-body diagram (FBD) is an essential tool in physics for visualizing the forces acting on a single object. It’s a simple drawing that depicts the object as a dot or a simple shape, with arrows representing all the forces applied to it. The direction of the arrow shows the direction of the force, and the length of the arrow is proportional to the force's magnitude.

In the case of the skater's hand, an FBD would show the centripetal force directed towards the center of the rotational path - this force is necessary to keep the hand moving in a circle. The diagram could also include gravitational force acting downwards, and perhaps a normal force if other surfaces were in contact. Understanding and drawing an accurate FBD is a foundational skill in solving problems related to forces and motion.
Rotational Motion and Centripetal Force
Rotational motion describes the movement of an object around a center or axis. An important aspect of rotational motion is the centripetal force, which is always directed toward the center of the rotational path and is responsible for changing the direction of an object's velocity. Without centripetal force, an object would move in a straight line due to inertia.

In the skater example, her hands are experiencing a centripetal force as they rotate around the axis of her body. This force is provided by the skater's wrist, ensuring her hands follow a circular path. The formula used, \(F = m\text{ω}^2r\), encapsulates the relationship between the mass \(m\), the angular velocity \(\text{ω}\), and the radius \(r\) of the circular motion. Understanding the interplay of these variables is critical in predicting and analyzing the motion of rotating systems in various physics scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You place a book of mass \(5.00 \mathrm{~kg}\) against a vertical wall. You apply a constant force \(\overrightarrow{\boldsymbol{F}}\) to the book, where \(F=96.0 \mathrm{~N}\) and the force is at an angle of \(60.0^{\circ}\) above the horizontal (Fig. \(\left.\mathbf{P} 5.75\right)\). The coefficient of kinetic friction between the book and the wall is \(0.300 .\) If the book is initially at rest, what is its speed after it has traveled \(0.400 \mathrm{~m}\) up the wall?

\(\mathrm{}\) A \(25.0 \mathrm{~kg}\) box of textbooks rests on a loading ramp that makes an angle \(\alpha\) with the horizontal. The coefficient of kinetic friction is \(0.25,\) and the coefficient of static friction is \(0.35 .(\) a) As \(\alpha\) is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid \(5.0 \mathrm{~m}\) along the loading ramp?

A box with mass \(m\) sits at the bottom of a long ramp that is sloped upward at an angle \(\alpha\) above the horizontal. You give the box a quick shove, and after it leaves your hands it is moving up the ramp with an initial speed \(v_{0}\). The box travels a distance \(d\) up the ramp and then slides back down. When it returns to its starting point, the speed of the box is half the speed it started with; it has speed \(v_{0} / 2 .\) What is the coefficient of kinetic friction between the box and the ramp? (Your answer should depend on only \(\alpha\).)

A \(1125 \mathrm{~kg}\) car and a \(2250 \mathrm{~kg}\) pickup truck approach a curve on a highway that has a radius of \(225 \mathrm{~m}\). (a) At what angle should the highway engineer bank this curve so that vehicles traveling at \(65.0 \mathrm{mi} / \mathrm{h}\) can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at \(65.0 \mathrm{mi} / \mathrm{h},\) find the normal force on each one due to the highway surface.

Jack sits in the chair of a Ferris wheel that is rotating at a constant \(0.100 \mathrm{rev} / \mathrm{s}\). As Jack passes through the highest point of his circular path, the upward force that the chair exerts on him is equal to one- fourth of his weight. What is the radius of the circle in which Jack travels? Treat him as a point mass.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.