If a 950 kg merry-go-round platform of radius 4.5 meters is driven by a mechanism located 2.0 meters from its center of rotation, how much force must the mechanism provide to get the platform moving at 5.5 revolutions per minute after 12 seconds if it were initially at rest

F = 213.75 N

Explanation:

First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.

ωf = ωi + αt

where,

ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s

ωi =initial angular velocity = 0 rad/s

t = time = 12 s

α = angular acceleration = ?

Therefore,

0.58 rad/s = 0 rad/s + α(12 s)

α = (0.58 rad/s)/(12 s)

α = 0.05 rad/s²

Now, we shall find the linear acceleration of the merry-go-round:

a = rα

where,

a = linear acceleration = ?

r = radius = 4.5 m

Therefore,

a = (4.5 m)(0.05 rad/s²)

a = 0.225 m/s²

Now, the force is given by Newton;s 2nd Law:

F = ma

where,

F = Force = ?

m = mass pf merry-go-round = 950 kg

Therefore,

F = (950 kg)(0.225 m/s²)

F = 213.75 N