Question

A 200-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 0.400 rev/s in 2.00 s

Answer 1

F = 187.5N.

Step-by-step explanation:

So, from the question above we are given the following parameters or data or information which is going to assist us in solving the question/problem;

=> Mass= 200-kg , => radius = 1.50 m, => angular speed of 0.400 rev/s, and time = 2.0 seconds.

Step one: the first step is to calculate or determine the angular speed. Here, the angular speed is Calculated in rad/sec.

Angular speed, w = 0.400 × 2π

= 2.51 rad/s.

Step two: determine the value of a.

Using the formula below;

W = Wo + a × time,t.

2.51 = 0 + a(2.0).

a= 1.25 rad/s^2.

Step three: determine constant force from the Torque.

Torque = I × a.

F = 1/2 × (200 kg) × 1.50 × 1.25.

F = 187.5N