A 165-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.) N

233.3N

Explanation:

Given:

radius 'r'= 1.5m

mass 'm'= 165kg

time 't'=2 sec

angular speed 'ω'= 0.6 rev/s

the magnitude of a torque is given by:

τ = F . r = I . α

where, 'α' is the angular acceleration and  'I' is the rotational inertia

F=  I . α/ r =>[ ( . m . )(2π . ω/t) ]/r

F=[( . 165 . )(2π . 0.6/2)]/1.5

F= 233.3N

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

spare some change and give it to the poor

explanation:

plz : ((