Question

a 105-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? (state the magnitude of the force.) toppr

Answer 1

To bring the merry-go-round from rest to an angular speed of 0.600 rev/s in 2.00 s, a force of 56.025 N needs 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, we need to apply a constant force on the rope. The force required can be calculated using the formula:

Force = (Moment of Inertia * Angular Speed) / Time

First, we need to find the moment of inertia of the merry-go-round disk, which is given as a uniform, solid, horizontal disk. The moment of inertia of a uniform disk is given by the formula:

Moment of Inertia = (1/2) * Mass * Radius^2

Substituting the given values, we have:

Moment of Inertia = (1/2) * 105 kg * (1.50 m)^2 = 186.75 kg · m²

Now, let's substitute the values into the force formula:

Force = (186.75 kg · m² * 0.600 rev/s) / 2.00 s = 56.025 N