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A solid uniform cylinder with mass 8.25 kg and diameter

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 61P Chapter 10

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 61P

A solid uniform cylinder with mass 8.25 kg and diameter 15.0 cm is spinning at 220 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.333. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?

Step-by-Step Solution:

Solution 61P Step 1 of 6: Given solid cylinder has mass M= 8.25 kg and diameter 15 cm, so the radius R=7.5cm= 0.075 m rotates with angular speed =220 rev/min and comes to rest =0 after 0 f undergoing = 5.25 rev angular displacement by the application of the brake on surface of cylinder with kinetic friction coefficient =0.333. Here need to calculate the normal k force that has to be applied to stop the cylinder. Step 2 of 6: Given data. Mass, M= 8.25 kg Radius, R=0.075 m kinetic friction coefficient =0.333 k Final angular speed, =0 f Initial angular speed, =220 rev/min 0 Using 1 rev =2 rad and 1 min= 60 sec = 220×2rad/s 0 60 =23.02 rad/sec 0 Angular displacement, = 5.25 rev Using 1 rev =2 rad = 5.25×2 rad =32.97 rad To find, Normal force , N= Step 3 of 6: In order to stop the rotating solid cylinder by the application of normal force, we need apply a normal force in such a way that the frictional force will be equal to the torque of the rotating cylinder. That is = N R k Where is torque, is coefficient of kinetic friction, N is normal force and R is radius. k Using = I I = kN R Solving for normal force, N= R …………...1 k Where I is moment of inertia and is angular acceleration.

Step 4 of 6

Chapter 10, Problem 61P is Solved
Step 5 of 6

Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

The full step-by-step solution to problem: 61P from chapter: 10 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: cylinder, brake, normal, Rim, Force. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. University Physics was written by and is associated to the ISBN: 9780321675460. Since the solution to 61P from 10 chapter was answered, more than 778 students have viewed the full step-by-step answer. The answer to “A solid uniform cylinder with mass 8.25 kg and diameter 15.0 cm is spinning at 220 rpm on a thin, frictionless axle that passes along the cylinder axis. You design a simple friction brake to stop the cylinder by pressing the brake against the outer rim with a normal force. The coefficient of kinetic friction between the brake and rim is 0.333. What must the applied normal force be to bring the cylinder to rest after it has turned through 5.25 revolutions?” is broken down into a number of easy to follow steps, and 82 words.

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