A child pushes a merry go-round from rest to a final angular speed of 0.6 rev/s with constant angular acceleration. In doing so, the child pushes the merry-go-round 1.8 revolutions. What is the angular acceleration of the merry go-round?
This is how i did this, but its wrong:
then, (0.6-0)/3=0.2 , so the answer
should be 0.2 rad/s^2, but its wrong I don't understand why!
Physics - Damon, Friday, March 23, 2012 at 6:07pm
v final = v initial + a t
.6 revs/s = a t
average speed during acceleration = .6/2 = .3 revs/s
1.8 revs = .3 revs/s * t
t = 6 seconds
a t = .6 revs/s
a = .6/6 = .1 revs/s^2
but 1 rev = 2 pi radians
a = .1 revs/s^2 * 2 pi radians/rev =
.2 pi radians/s^2
= .628 radians/s^2
Physics - Adam, Friday, March 23, 2012 at 7:50pm
Thanks :) Damon!