Question

# A child of mass m=15kg stands on a merry-go-round. The radius of the merry-go-round is R=2m...

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?

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 0o 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 90o 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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