/*! 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} Q.20 A roller coaster car crosses the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. What is the ratio of the normal force to the gravitational force?

Short Answer

Expert verified

The ratio of the normal force to the gravitational force is 3.

Step by step solution

01

Given information

Given in the question that, A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.

02

Explanation

From the information we observed that mV2ris the result of Nand mg

Here,

m-The mass of the roller coaster

g-The acceleration due to gravity

r-The radius of the circular loop

V- The velocity

Therefore,

mV2r+N=mg

N=mg-mV2r

V2=4gr

Since roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.

Therefore,

N=mg-m(4gr)r

N=-3mg

Hence, the ratio of the normal force to the gravitational force is3

03

Final answer

The ratio of the normal force to the gravitational force is 3.

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!

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

A golfer starts with the club over her head and swings it to reach maximum speed as it contacts the ball. Halfway through her swing, when the golf club is parallel to the ground, does the acceleration vector of the club head point (a) straight down, (b) parallel to the ground, approximately toward the golfer’s shoulders, (c) approximately toward the golfer’s feet, or (d) toward a point above the golfer’s head? Explain.

Space scientists have a large test chamber from which all the air can be evacuated and in which they can create a horizontal uniform electric field. The electric field exerts a constant horizontal force on a charged object. A 15gcharged projectile is launched with a speed of 6.0m/s at an angle35° above the horizontal. It lands 2.9m in front of the launcher. What is the magnitude of the electric force on the projectile?

The father of Example 8.2 stands at the summit of a conical hill as he spins his 20 kg child around on a 5.0 kg cart with a 2.0-m-long rope. The sides of the hill are inclined at 20o­. He
again keeps the rope parallel to the ground, and friction is negligible. What rope tension will allow the cart to spin with the same 14 rpm it had in the example?

A 500 g steel block rotates on a steel table while attached to a 2.0-m-long massless rod. Compressed air fed through the rod is ejected from a nozzle on the back of the block, exerting a thrust force of 3.5 N. The nozzle is 70o­ from the radial line, as shown in FIGURE P8.62. The block starts from rest.
a. What is the block’s angular velocity after 10 rev?
b. What is the tension in the rod after 10 rev?

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in FIGURE P8.51.
a. Suppose the ring rotates once every 4.5 s. If a rider’s mass is 55 kg, with how much force does the ring push on her at the top of the ride? At the bottom?
b. What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

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.