Conservation of momentum on merry-go-round

[tandoorichicken]

A playground merry-go-round has a disk-shaped platform that rotates with negligible friction about a vertical axis. The disk has a mass of 200kg and a radius of 1.8 m. A 36-kg child rides at the center of the merry-g-round while a playmate sets it turning at 0.25 rev/sec. If the child then walks along a radius to the outer edge of the disk, how fast will the disk be turning.

Work:
[tex]\omega_0 = 0.5\pi [/tex] rad/sec.
[tex] I_0 = \frac{1}{2} m r^2 [/tex] (MOI for disk).
[tex] L = I_0\omega_0 = \frac{1}{2} m r^2 \omega_0 [/tex]
So, if angular momentum is conserved, [tex] L = I_f\omega_f [/tex]
My only problem is figuring out the MOI for a disk with an object spinning around the outer edge. Anyone know?

 

[Doc Al]

Originally posted by tandoorichicken
My only problem is figuring out the MOI for a disk with an object spinning around the outer edge. Anyone know?

Treat the child as a point mass. Its moment of inertia is [itex]mr^2[/itex]. The total MOI is that of the disk plus that of the child.

 

[M98Ranger]

The moment of inertia (MOI) for a disk with an object spinning around the outer edge can be calculated using the parallel axis theorem. This states that the MOI of a body about an axis parallel to its center of mass is equal to the MOI of the body about its center of mass plus the product of its mass and the square of the distance between the two axes. In this case, the MOI for the disk with the child at the outer edge can be calculated as:

I_f = I_0 + md^2

Where:
I_f = MOI of the disk with the child at the outer edge
I_0 = MOI of the disk with the child at the center
m = mass of the child
d = distance between the center of the disk and the outer edge (equal to the radius of the disk)

Substituting the known values into the equation, we get:

I_f = \frac{1}{2} (200 kg)(1.8 m)^2 + (36 kg)(1.8 m)^2 = 388.8 kgm^2

Now, using the conservation of angular momentum equation, we can solve for the final angular velocity of the disk:

L = I_f\omega_f
\frac{1}{2} m r^2 \omega_0 = 388.8 kgm^2 \omega_f
\omega_f = \frac{1}{2} m r^2 \omega_0 \div 388.8 kgm^2
\omega_f = 0.5\pi rad/sec \div 388.8 kgm^2
\omega_f = 0.00128 rad/sec

Therefore, the disk will be turning at a final angular velocity of 0.00128 rad/sec after the child walks to the outer edge. This is a very small change compared to the initial angular velocity of 0.5π rad/sec, as the child's mass is relatively small compared to the mass of the disk. This demonstrates the conservation of angular momentum, where the total angular momentum of the system remains constant despite changes in the distribution of mass.