Question

A 145-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 145-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 would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s? (State the magnitude of the force.)
 N

Answer

Given:

Mass, m=145 kg

Disc radius, r=1.5 m

Angular speed=0.6 rev/s

t=2 s

Calculating the angular speed in rad/s:

The angular speed in rad/s is given by,

ω=0.6×2π

On solving,

ω=3.768 rad/s

 

Using the expression,

ω=ω0+αt

On substituting the values,

3.768 rad/s=0+α(2 s)  ω0=0

On solving,

α=1.884 rad/s2

The torque is given by,

τ=

F×r=12mr2α

On simplifying,

F=12mrα

 

On substituting the values,

F=12×145 kg×1.5 m×1.884 rad/s2

On solving,

F=204.885 N

 

Hence, the required force is 204.885 N.

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