Question

A 245-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 magnitu

Answer 1

F = 345.45 N

Step-by-step explanation:

Angular acceleration of the disc is given as rate of change in angular speed

it is given by formula

`\alpha = (d\omega)/(dt)`

`\alpha = (2\pi(0.600))/(2)`

`\alpha = 1.88 rad/s^2`

now we know that moment of inertia of the solid uniform disc is given as

`I = (1)/(2)mR^2`

`I = (1)/(2)245(1.50)^2`

`I = 275.625 kg m^2`

now we have an equation for torque as

`\Tau = I\alpha`

`r F = 275.625(1.88)`

`F = (275.625(1.88))/(1.50)`

`F = 345.45 N`