Question

A 300 kg merry-go-round in the shape of a horizontal disk with a radius of 1.9 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 3.6 rad/s in 3.5 s

Answer 1

Answer:

557.7Nm

Explanation:

We are given that

$m=300 kg$

Radius,r=1.9 m

Angular speed,$\omega=3.6rad/s$

Time,t=3.5 s

Initial angular velocity,$\omega_0=0$

We have to find the large torque would have to be exerted .

$\alpha=\frac{\omega-\omega_0}{t}=\frac{3.6-0}{3.5}=1.03rad/s^2$

$\tau=I\alpha$

$\tau=\frac{1}{2}mr^2\alpha$

Where $I=\frac{1}{2}mr^2$

Substitute the values

$\tau=\frac{1}{2}(300)(1.9)^2(1.03)=557.7Nm$