Consider the angular moment of inertia to be I. Then the moment of inertia due to m at some r is 1/2 m r^2
conservation of momentum:
Iinitial wi= Ifinal * Wf
(I+1/2mr^2)w=(I + 1/2 m 0^2)wf
(I+1/2 45*2).5=(I wf)
you are given I as 360
405*.5=360 wf
wf=.563 rad/sec
check me.
A merry-go-round is rotating at 4.775 revs/min with a 45 kg child on the outer edge. The radius is 2.0 m. The child moves to the centre of the merry go round. What is the final angular velocity of the merry go round. The moment of inertia of the merry-go-round is 360kgm^s
First thing, w initial=0.5 rad/s
So L initial= 90kgm^2/s
I'm going to use conservation of momentum to find the final angular velocity, but I'm not sure what the final momentum of inertia is (ie. when the child moves to the centre?)
I tried taking 360 and adding it to 1/2MR^2 (to find the contribution of the moment of inertia of the child) trying to account for the child being at the centre.. then using conservation of momentum and solving for Wf.. but i didn't get the correct answer
2 answers
Yes!
That's right!
Except the moment of inertia of the person is just mr^2 (not 1/2mr^2), making the final answer 0.75, aka the correct answer in the solutions manual!
Thanks so much :)
That's right!
Except the moment of inertia of the person is just mr^2 (not 1/2mr^2), making the final answer 0.75, aka the correct answer in the solutions manual!
Thanks so much :)
2 answers