Question

A child of mass m=15kg stands on a merry-go-round. The radius of the merry-go-round is R=2m and the mass is M=75kg. The merry-go-round is considered to be a solid disk. The merry-go round initially at rest and the child stands near the edge. What will the angular velocity of the merry-go-round be if the child jumps off

a.) In the radial direction with a velocity of +4m/s?

b.) perpindicular to the radial direction with a velocity of +4m/s?

c.) At an angle of 45 degrees above the radial direction with a velocity of 4 m/s?

Answer #1

a] Use conservation of angular momentum

Initial angular momentum is zero.

so,

if the child jumps radially, the angle between the radius vector
and the velocity vector is 0^{o} and so the final angular
momentum of the merry-go-round is also zero. Therefore, the angular
velocity of the merry-go-round after the jump is still zero.

b]

Here, the angle between r and v is 90^{o} since the jump
is tangential.

=>

=>

=>

this is the angular velocity of the Merry-go-round after the child jump's out tangentially.

c]

.

this is the angular velocity of the Merry-go-round after the child jump's out at an angle of 45 degrees from radial direction.

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