Question

A 312 kg merry-go-round in the shape of a horizontal disk with a radius of 1.1 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.3 rad/s in 3.6 s?

Answer 1

Answer:172.90 N-m

Step-by-step explanation:

Given

mass of merry-go-round
`m=312\ kg`

radius of disk
`r=1.1\ m`

Final angular speed
`\omega _f=3.3\ rad/s`

time
`t=3.6\ s`

initial angular speed
`\omega _0=0\ rad/s`

using
`\omega _f=\omega _0+\alpha \cdot t`

where
`\omega _f`=final angular speed

`\omega _i`=Initial angular speed

`\alpha`=angular speed

`t`=time

`3.3=0+\alpha * 3.6`

`\alpha =0.916\ rad/s^2`

Torque can be written by

`T=I* \alpha`

where I=moment of Inertia

`T=(mR^2)/(2)* \alpha`

`T=(312* 1.1^2)/(2)* 0.916`

`T=172.90\ N-m`