The magnitude of the force required for the Merry-go-round is 60.93 N.
The mass of the Merry-Go-Round is 190 kg the Merry-Go-Round is a uniform solid horizontal disc of radius 1.5 m.
The Merry-Go-Round is rotating by wrapping a rope around the rim of the disc.
It is required to bring the Merry-Go-round from rest to an angular speed of 0.7 Rev/s in 2 seconds.
Here, we can use the relation,
W' = W + at
Where,
W' is final angular speed,
W is initial angular speed,
a is angular acceleration and,
t is the time required.
Putting values,
0.4396 = a(2)
a = 0.2198 rad/s².
The relation between angular and linear acceleration is,
A = ra
A is linear acceleration,
A = 0.3207 m/s².
Force needed to be applied on the rope,
F = MA
F = 190 x 0.3207
F = 60.93 N.
So, the magnitude of the required force is 60.93N.
To know more about acceleration, visit,
https://brainly.com/question/605631
#SPJ4
