A 175 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?

Question

I first solved for the angular acceleration and got 0.2 rev/s^2

Then I used 1/2mr*(angular acceleration)
and got 26.25 N, but that answers is incorrect.

Please help, thanks!

Answer 1

Let "omega" = w = angular velocity
1 rev/s = 2pi rad/s
angular velocity = w = (0.400)(2pi rad/s) = 2.5133 rad/s
Linear velocity = v = wr
v = (2.5133 rad/s)(1.5m/rad) = 3.77 m/s
Linear accel. = (change in velocity)/time = a
a = (Vf - Vi)/t = (3.77 m/s - 0)/2.00 s = ______?
Force = mass x acceleration = ____?
(use mass given and the value of a)

Answer 2

329.9

Answer 3

neither one of these are correct