What is the NEW angular speed? Steps to answer ?

Question

A merry-go-round rotates at the rate of 0.15 rev/s with an 85-kg man standing at a point 2.0 m from the axis of rotation.

QUESTION: What is the new angular speed when the man walks to a point 0.7 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.
In Rad/Sec?

Answer 1

This is a classic conservation of angular momentum question.
We will assume frictionless motion.
The angular momentum is always:

1/2 r^2 m for the disk
plus
d^2 m for the man
the sum multiplied by the angular speed

since the momentum is conserved:

(1/2*4*25+85*4)*.15=
(1/2*4*25+85*.7^2) speed

speed=.64 rev/s

I'll let you convert it to radians per second

j

Answer 2

As acceleration is consistent, to acquire the entire type of rev for the hollow era you should use the favourite angular speed. 3,2 rev/s for both activities ( starting up and preventing). which will bring about a finished time of 16.2 s x 3,2 rev/s= fifty one,80 4 rev