Question
A merry-go-round accelerating uniformly from rest achieves its operating speed of 3.4 rpm in 2 revolutions. What is the magnitude of the angular acceleration?
Answers
Damon
4 pi radians = distance gone
3.4 *2 pi/60 = final speed in radians/s
average speed = half of final = 3.4 pi/60
so
4 pi = (3.4 pi/60)t
so
t = 4*60/3.4
v = Vi + a t
3.4*2 pi/60 = a (4*60/3.4)
a = (3.4^2 /6^2)(pi/2)
3.4 *2 pi/60 = final speed in radians/s
average speed = half of final = 3.4 pi/60
so
4 pi = (3.4 pi/60)t
so
t = 4*60/3.4
v = Vi + a t
3.4*2 pi/60 = a (4*60/3.4)
a = (3.4^2 /6^2)(pi/2)
Shamsuddeen
The magnitud of angular 6.8