/*! 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 21 A \(55.0-\mathrm{kg}\) ice skate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 21

A \(55.0-\mathrm{kg}\) ice skater is moving at \(4.00 \mathrm{~m} / \mathrm{s}\) when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius \(0.800 \mathrm{~m}\) around the pole. (a) Determine the force exerted by the horizontal rope on her arms. (b) Compare this force with her weight.

Short Answer

Expert verified
The force exerted by the horizontal rope on her arms is about 1100 N, which is far greater than her weight (about 539 N).

Step by step solution

01

Find the Centripetal Force

When an object moves in a circle, it experiences a force towards the center of the circle. This force is known as the centripetal force. It can be calculated using the formula \( F = m \cdot v^2 / r \), where \( F \) is the centripetal force, \( m \) is the mass of the object, \( v \) is the velocity of the object, and \( r \) is the radius of the circle. Fill in the given values to calculate the force. \( F = 55.0 \cdot (4.00)^2 / 0.800 \) kg.
02

Calculate the force

Now that the values have been filled in, the calculation can be made. After performing the calculation, we find that \( F \approx 1100 \) Newtons.
03

Find the skater's weight

The skater's weight can be determined using the equation for force due to gravity \( F = m \cdot g \), where \( g \) is the gravitational acceleration (approximately \( 9.8 m/s^2 \) on Earth). Substituting the given values, we find that the weight of the skater is \( 55.0 kg \cdot 9.8 m/s^2 \approx 539 N \).
04

Compare the two forces

Now the centripetal force exerted by the horizontal rope on her arms (calculated in step 2) can be compared to her weight (calculated in step 3). The centripetal force is much larger than her 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 91影视!

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

An electric motor rotating a workshop grinding wheel at a rate of \(1.00 \times 10^{2} \mathrm{rev} / \mathrm{min}\) is switched off. Assume the wheel has a constant negative angular acceleration of magnitude \(2.00 \mathrm{rad} / \mathrm{s}^{2}\). (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in part (a)?

Show that the escape speed from the surface of a planet of uniform density is directly proportional to the radius of the planet.

A \(40.0\)-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of \(18.0 \mathrm{~m}\). (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?

A stuntman whose mass is \(70 \mathrm{~kg}\) swings from the end of a \(4.0-\mathrm{m}-\mathrm{long}\) rope along the arc of a vertical circle. Assuming he starts from rest when the rope is horizontal, find the tensions in the rope that are required to make him follow his circular path (a) at the beginning of his motion, (b) at a height of \(1.5 \mathrm{~m}\) above the bottom of the circular arc, and (c) at the bottom of the arc.

A coin rests \(15.0 \mathrm{~cm}\) from the center of a turntable. The coefficient of static friction between the coin and turntable surface is \(0.350\). The turntable starts from rest at \(t=0\) and rotates with a constant angular acceleration of \(0.730 \mathrm{rad} / \mathrm{s}^{2}\). (a) Once the turntable starts to rotate, what force causes the centripetal acceleration when the coin is stationary relative to the turntable? Under what condition does the coin begin to move relative to the turntable? (b) After what period of time will the coin start to slip on the turntable?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Electromagnetism

Read Explanation

Quantum Physics

Read Explanation

Translational Dynamics

Read Explanation

Scientific Method Physics

Read Explanation

Torque and Rotational Motion

Read Explanation
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.