   Chapter 8, Problem 45P

Chapter
Section
Textbook Problem

A 150.-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?

To determine
The force exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500rev/s in 2.00s .

Explanation

Given info: The mass of the merry-go-round is 150kg , the radius of the merry-go-round is 1.50m , final angular speed of the merry-go-round is 0.500rev/s , and the time for the merry-go-round to reach an angular speed is 2.00s .

Explanation: The angular acceleration of the merry-go-round is defined as α=(ωfωi)/Δt with zero initial angular velocity and its moment of inertia is written as I=MR2/2 . The torque of the merry-go-round is τ=rF=Iα and from these expressions, the force exerted on the rope is determined.

The formula for the force exerted on the rope to bring the merry-go-round from rest to an angular speed is,

F=(MR2/2)(ωfΔt)R

• M is mass of the merry-go-round.
• R is radius of the merry-go-round.
• ωf is final angular speed of the merry-go-round.
• Δt is the time for the merry-go-round to reach an angular speed.

Substitute 150kg for M , 1

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