A 180-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

The force acting on the rope is 211.95 N.

Explanation:

Given that,

Mass of merry-go-round, m = 180 kg

Radius of merry-go-round, r = 1.5 m

Initial angular speed is 0 it was at rest

Final angular speed,

Time, t = 2 s

When it moves in circular path, the torque acting on it is given by :

So, the force acting on the rope is 211.95 N.

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

inverse v=i*r

nice nice