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

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

$3.3=0+\alpha \times 3.6$

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

Torque can be written by

$T=I\times \alpha$

where I=moment of Inertia

$T=\frac{mR^2}{2}\times \alpha$

$T=\frac{312\times 1.1^2}{2}\times 0.916$

$T=172.90\ N-m$