Question

g A 240-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 0.600 rev/s in 2.00 s

Answer 1

Answer:

339.3 N

Explanation:

First, we start by converting the units.

1 rev/s = 2π rad/s, so

0.6 rev/s = 2π * 0.6 rad/s

0.6 rev/s = 1.2π rad/s

0.6 rev/s = 3.77 rad/s

Now we apply the equation of motion,

W(f) = w(o) + αt

3.77 = 0 + α * 2

3.77 = 2α

α = 3.77/2

α = 1.885 rad/s²

Torque = I * α

Torque = F * r

This means that

I * α = F * r, where I = 1/2mr²

Substituting for I, we have

1/2mr²α = F * r, making F the subject of formula, we have

F = 1/2mrα, then we substitute for the values

F = 1/2 * 240 * 1.5 * 1.885

F = 678.6 / 2

F = 339.3 N