Answer:
a) [tex]1209.6 kg m^2/s[/tex]
b) [tex]302.4 kg m^2/s[/tex]
Explanation:
a)
The angular momentum of an object in circular motion is given by the formula
[tex]L=m\omega r^2[/tex]
where
m is its mass
[tex]\omega[/tex] 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
[tex]f=\frac{1}{5}=0.2 Hz[/tex]
And so his angular speed is
[tex]\omega=2\pi f=2\pi 0.2=1.26rad/s[/tex]
Therefore, the angular momentum is
[tex]L=(60)(1.26)(4)^2=1209.6 kg m^2/s[/tex]
b)
For a child sitting halfway of the merry-go-around, his distance from the axis of rotation is
[tex]r=\frac{4 m}{2}=2 m[/tex]
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,
[tex]\omega=1.26 rad/s[/tex]
Therefore, the angular momentum here is:
[tex]L=m\omega r^2 = (60)(1.26)(2)^2=302.4 kg m^2/s[/tex]
