   Chapter 8, Problem 49P

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

Explanation: The angular acceleration of the cylindrical shaped merry-go-round is α=τ/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=Iω2/2=MR2ω2/4 .

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

α=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 50.0N for F , 9.80m/s2 for g , 800N for W , 1.50m for R to find α .

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

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 