/*! 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. 48 The 10 mg bead in FIGURE P8.48 i... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Q. 48 (page 201)

The 10 mg bead in FIGURE P8.48 is free to slide on a frictionless wire loop. The loop rotates about a vertical axis with angular velocity 蝇. If 蝇 is less than some critical value 蝇c the bead sits at the bottom of the spinning loop. When 蝇 > 蝇c the bead moves out to some angle 胃

  1. What is 蝇c in rpm for the loop shown in the figure?
  2. At what value of 蝇c in rpm is 胃=30o

Short Answer

Expert verified

1) Angular Speed of the ball is 蝇c=gRcos胃

2) The angular speed 蝇c in rpm is 143.52 rpm.

Step by step solution

01

Part(1) Step 1 : Given Information

R = 5cm =0.05m

胃=30o

02

Part(1) Step 2: Explanation

First draw the Free body diagram and equate forces components.

Ncos胃=mg.............(1)

and

Nsin胃=m蝇2r...................(2)

Divide (2) by (1) we get

tan胃=蝇2rgSubstituter=Rsin胃tan胃=蝇2Rsin胃g

Solve it for 蝇, we get

蝇=gRcos胃...............................(3)

03

Part(2) Step 1 : Given Information

R = 5cm=0.05m

胃=30o

04

Part(2) Step 2 : Explanation

Substitute the values in the equation (2), we get

蝇=(9.8m/s2)(0.05m脳cos300)=15.03rad/sec

Convert in rpm

蝇=(15.03rad/sec)(1rpm2蟺rad)(60sec1min)=143.52rpm

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

In the Bohr model of the hydrogen atom, an electron (mass m=9.1脳10-31kg)orbits a proton at a distance of 5.3脳10-11m. The proton pulls on the electron with an electric force of8.2脳10-8N. How many revolutions per second does the electron make?

A satellite orbiting the moon very near the surface has a period of 110min. What is free-fall acceleration on the surface of the moon?

A 1500kg car drives around a flat 200-m-diameter circular track at 25m/s. What are the magnitude and direction of the net force on the car? What causes this force?

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鈥檚 shoulders, (c) approximately toward the golfer鈥檚 feet, or (d) toward a point above the golfer鈥檚 head? Explain.

A 2.0 kg ball swings in a vertical circle on the end of an 80-cm-long string. The tension in the string is 20 N when its angle from the highest point on the circle is 胃 = 30颅o.
a. What is the ball鈥檚 speed when 胃 = 30o颅?
b. What are the magnitude and direction of the ball鈥檚 acceleration when 胃 = 30o颅?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Classical Mechanics

Read Explanation

Famous Physicists

Read Explanation

Fluids

Read Explanation

Particle Model of Matter

Read Explanation

Medical Physics

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.