Bob weights 36Kg sees a merry-go-round which is initially not moving. As he jump
ID: 1563636 • Letter: B
Question
Bob weights 36Kg sees a merry-go-round which is initially not moving. As he jumps on it immediately before making contact with it he has horizontal velocity of 2 m/s that is also tangential to the outside of the merry-go-round (perpendicular to its radius). He lands a distance 3 m from the center and holds on so that he rotates together with the merry-g-r. If the merry-g-r has a moment of inertia of 951 kg*m^2 and Bob can be treated as point mass what is the angular speed of the merry-go-round after Bob lands on it? Friction and air resistance neglected.
Thank you!
Explanation / Answer
initial angularmomentum= Li= m vr=36(2)(3)=216 kgm2 /s
moment of inertia ofmerry g-r=951 kgm2
moment of inertia of Bob= mr2 =36(3)2.=324 kgm2
total moment of inertia=951+324=1275 kgm2
let w= final angular velocity
final angular momentum= Lf= i w= 1275w
now according to law of conservatio of angular momentum, Lf=Li
1275w=216
w=0.169 rad/sec