A125-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 0.700 rev/s in 2.00 s? (state the magnitude of the force.)

201.6 N

Explanation:

m = mass of disk shaped merry-go-round = 125 kg

r = radius of the disk = 1.50 m

w₀ = Initial angular speed = 0 rad/s

w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s

t = time interval = 2 s

α = Angular acceleration

Using the equation

w = w₀ + α t

4.296 = 0 + 2α

α = 2.15 rad/s²

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²

F = constant force applied

Torque equation for the merry-go-round is given as

r F = I α

(1.50) F = (140.625) (2.15)

F = 201.6 N

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .