Question

a 60 kg child sits on on the outer edge a playground merry-go-around with radius of 4 meters that spins around at 1 revolution every 5 seconds, mercilessly bringing the child ever closer to nausea. What is the child's angular momentum? ( Remember to include the units.) What is the angular momentum of another child with the same mass who is sitting half way out from the center?

Answer 1

Answer:

a) $1209.6 kg m^2/s$

b) $302.4 kg m^2/s$

Explanation:

a)

The angular momentum of an object in circular motion is given by the formula

$L=m\omega r^2$

where

m is its mass

$\omega$ is its angular velocity

r is the distance of the object from the axis of rotation

For the child in this problem, we have:

m = 60 kg is his mass

r = 4 m is the radius of the merry-go-around

Here the child completes 1 revolution every 5 seconds; so his frequency is

$f=\frac{1}{5}=0.2 Hz$

And so his angular speed is

$\omega=2\pi f=2\pi 0.2=1.26rad/s$

Therefore, the angular momentum is

$L=(60)(1.26)(4)^2=1209.6 kg m^2/s$

b)

For a child sitting halfway of the merry-go-around, his distance from the axis of rotation is

$r=\frac{4 m}{2}=2 m$

The mass of the child is the same as before,

m = 60 kg

And the angular speed is the  same as well: in fact, the merry go around is a rigid body, so all its points cover the same angle in the  same time; so, they all have the same angular speed. Therefore,

$\omega=1.26 rad/s$

Therefore, the angular momentum here is:

$L=m\omega r^2 = (60)(1.26)(2)^2=302.4 kg m^2/s$