Question

State the law of conservation of angular

momentum.

A merry-go-round has moment of inertia

4500 kg m2. It is mounted on a frictionless

vertical axle. It is initially rotating at an

angular speed of 1 rpm. A girl jumps onto the

merry-go-round in the radial direction. If the

moment of inertia of the girl is 500 kg m 2, find

the angular speed of rotation of the merry-go-

round.​

Answer 1

Answer:

The merry-go-round has a moment of inertia

I

. When the child hops it on and sits on the edge the moment of inertia changes to

I

′

=

I

+

m

r

2

where

m

is mass of child and

r

radius of merry-go-round.

As the child hops onto the edge there is no external force acting on the system. As such angular momentum

L

is conserved. Therefore we have

L

′

=

L

Let

ω

′

be changed angular velocity. Using the definition of angular momentum we have

I

′

ω

′

=

I

ω

⇒

ω

′

=

I

ω

I

′

Inserting given values we get

ω

′

=

290

×

11

290

+

30

×

(

1.3

)

2

ω

′

=

9.4

rpm