Question

A 416 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 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 3.7 rad/s in 2.9 s

Answer 1

The torque exerted on the merry-go-round is 766.95 Nm

Step-by-step explanation:

Given;

mass of the merry-go-round, m = 416 kg

radius of the disk, r = 1.7 m

angular speed of the merry-go-round, ω = 3.7 rad/s

time of motion, t = 2.9 s

The torque exerted on the merry-go-round is calculated as;

`\tau = Fr= I\alpha\\\\\tau = ((1)/(2) m r^2)((\omega )/(t) )\\\\\tau = ((1)/(2) * 416 * 1.7^2)( (3.7)/(2.9) )\\\\\tau = 766.95 \ Nm`

Therefore, the torque exerted on the merry-go-round is 766.95 Nm