/*! 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 A 52-kg ice skater spins about a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 52-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 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
Centripetal force is 40.84 N and is about 6.41 times the hand's weight.

Step by step solution

01

Understand the Problem

Begin by understanding that the skater is spinning around a vertical axis with her arms outstretched. Each hand makes a circular path, requiring a centripetal force provided by her wrists.
02

Calculate Mass of One Hand

Determine the mass of one hand using 1.25% of the skater's total body weight. \[ m_{hand} = 0.0125 \times 52\, \text{kg} = 0.65\, \text{kg} \]
03

Calculate the Radius of Circular Motion

Since the total span from one hand to the other is 1.50 m, the radius from the axis to one hand is half of 1.50 m. \[ r = \frac{1.50}{2} = 0.75\, \text{m} \]
04

Determine Centripetal Force Requirement

The centripetal force for one hand is calculated using the formula for centripetal force: \[ F_c = m_{hand} \cdot r \cdot \omega^2 \] where \( \omega = 2\pi \times 2 \) (since the skater makes 2 revolutions per second), so \( \omega = 4\pi \) Substitute values to find: \[ F_c = 0.65 \cdot 0.75 \cdot (4\pi)^2 \approx 40.84\, \text{N} \]
05

Express Force as a Multiple of the Weight of Hand

The weight of one hand \( W_{hand} \) is given by: \[ W_{hand} = m_{hand} \cdot g = 0.65 \times 9.8 = 6.37\, \text{N} \] Now express the centripetal force as a multiple of the hand's weight:\[ \text{Multiple} = \frac{F_c}{W_{hand}} = \frac{40.84}{6.37} \approx 6.41 \]
06

Draw Free-Body Diagram

For part (a), draw a simple diagram showing a hand with two force vectors: the force from the wrist acting inward (centripetal force) and the hand's weight acting downward (though the weight does not affect horizontal motion).

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.

Circular Motion
Circular motion refers to the movement of an object along the circumference of a circle. This is what happens when you swing an object around a central point, like the hands of a skater spinning around their body. In this context, each hand of the ice skater moves in a circular path around the vertical axis passing through the skater's body.

As the skater spins, each hand requires a force to keep moving in that circular path. This is known as the centripetal force, and it must always act towards the center of the circular path. The lack of any horizontal force would cause the hands to move in a straight line due to inertia, flying outward from the body like a frisbee.

This force is provided by the muscles in the skater's wrist. The muscles pull the hands inward, allowing them to stay in circular motion at the radius determined by the distance from her body.
Free-Body Diagram
A free-body diagram is a visual tool used to show all the forces acting on an object. It helps in analyzing the forces to solve physics problems, such as determining the net force and the resulting motion of an object.

In the skater's problem, we draw a free-body diagram for just one hand. The hand experiences two main forces:
  • An inward force, which is the centripetal force provided by the wrist. This force keeps the hand moving in a circle.
  • The gravitational force, otherwise known as the weight of the hand, pulls downward.
The centripetal force vector should be directed horizontally towards the center of the circle (the skater's body), while the weight vector should be pointing downward due to gravity. This diagram helps us visualize and compute the necessary forces for the skater's hand to remain in circular motion.
Rotational Motion
Rotational motion involves objects that rotate around a central point or axis. In this situation, the ice skater exhibits rotational motion as she spins around her vertical axis. This type of motion is described by the angular velocity, which in this problem, is given as the number of turns she makes per second.

The angular velocity, denoted as \(\omega\), is a crucial aspect of rotational motion. It's calculated by converting revolutions per second into radians per second, utilizing the formula \(\omega = 2\pi \times \,\text{turns per second}\). In our case, because the skater makes two turns per second, \(\omega = 4\pi\,\text{rad/s}\).

This angular speed determines how fast each hand moves along the circular path when the skater spins. Together with the radius of the hand's path, it helps calculate the centripetal force necessary to maintain the hand's circular path, thus establishing the interplay between circular and rotational motions in maintaining equilibrium.

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

Two crates connected by a rope lie on a horizontal surface (\(\textbf{Fig. E5.37}\)). Crate \(A\) has mass \(m_A\), and crate \(B\) has mass \(m_B\). The coefficient of kinetic friction between each crate and the surface is \(\mu_k\). The crates are pulled to the right at constant velocity by a horizontal force \(F\). Draw one or more free-body diagrams to calculate the following in terms of \(m_A\), \(m_B\), and \(\mu_k\): (a) the magnitude of \(F\) and (b) the tension in the rope connecting the blocks.

A box with mass m is dragged across a level floor with coefficient of kinetic friction \(\mu_k\) by a rope that is pulled upward at an angle \(\theta\) above the horizontal with a force of magnitude \(F\). (a) In terms of \(m, \mu_k, \theta\), and \(g\), obtain an expression for the magnitude of the force required to move the box with constant speed. (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90-kg patient across a floor at constant speed by pulling on him at an angle of 25\(^\circ\) above the horizontal. By dragging weights wrapped in an old pair of pants down the hall with a spring balance, you find that \(\mu_k\) = 0.35. Use the result of part (a) to answer the instructor's question.

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box's acceleration?

A hammer is hanging by a light rope from the ceiling of a bus. The ceiling is parallel to the roadway. The bus is traveling in a straight line on a horizontal street. You observe that the hammer hangs at rest with respect to the bus when the angle between the rope and the ceiling of the bus is 56.0\(^\circ\). What is the acceleration of the bus?

A pickup truck is carrying a toolbox, but the rear gate of the truck is missing. The toolbox will slide out if it is set moving. The coefficients of kinetic friction and static friction between the box and the level bed of the truck are 0.355 and 0.650, respectively. Starting from rest, what is the shortest time this truck could accelerate uniformly to 30.0 m/s without causing the box to slide? Draw a free-body diagram of the toolbox.

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.