Chapter 8, Problem 49P

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300

Chapter
Section

### College Physics

11th Edition
Raymond A. Serway + 1 other
ISBN: 9781305952300
Textbook Problem

# A horizontal 800.-N merry-go-round of radius 1.50 m is started from rest by a constant horizontal force of 50.0 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 3.00 s. (Assume it is a solid cylinder.)

To determine
The kinetic energy of the merry-go-round after 3.00s .

Explanation

Given info: The radius of the merry-go-round is 1.50â€‰m , its weight is 800â€‰N , the constant horizontal force is 50.0â€‰N , and the acceleration due to gravity is 9.80â€‰m/s2 , initial angular speed is 0â€‰rad/s , the angular acceleration of the merry-go-round is 0.817â€‰rad/s2 , the time for the merry-go-round to reach such final angular speed is 3.00â€‰s , final angular speed of the merry-go-round is 2.45â€‰rad/s , radius of the merry-go-round is 1.50â€‰m , its weight is 800â€‰N and the acceleration due to gravity is 9.80â€‰m/s2 .

Explanation: The angular acceleration of the cylindrical shaped merry-go-round is Î±=Ï„/I=RF/(MR2/2)=2F/MR . For the mass of the merry-go-round, M=W/g . The kinetic energy of the merry-go-round is KEr=IÏ‰2/2=MR2Ï‰2/4 .

The formula for the angular acceleration of the merry-go-round,

Î±=2FgWR

• F is constant horizontal force.
• g is acceleration due to gravity.
• W is weight of the merry-go-round.
• R is merry-go-round.

Substitute 50.0â€‰N for F , 9.80â€‰m/s2 for g , 800â€‰N for W , 1.50â€‰m for R to find Î± .

Thus, the angular acceleration of the merry-go-round is 0.817â€‰rad/s2

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