Question

I need help with # 2. Finding the moment of inertia when the arms are up and when the arms are down. I have tried and my number is to small. Question is linked http://imgur.com/zcyzk

Answer 1

throughout the sum i used m.o.i=mr^2 formula
first when arms are stretched we have mass 2.8 kg and distance from axis of rotation as 0.23+(0.65/2)=0.555m
(2.8*0.555*0.555*2)+3 =4.72494 KGM^2 gives the m.o.i for first case(2 arms)

next we have distance from axis of rotation as shown in diagram
2.8*0.28*0.28=0.21592
2( 0.21592)+3=3.43904KGM^2 m.o.i for 2nd case

as momentum is conserved
I.W=I'.W'
4RAD/S*4.72494 KGM^2=W rad/s*3.43904KGM
finally
omega(w')=5.4956 rad/s