Question

A Lazy Susan is a circular disk that can spin on a center axis. Imagine a bowl of fruit on the disk as in the photo below. The bowl has a mass of 4 kg and the Lazy Susan has a mass of 5.3 kg. If you take the bowl that is sitting on the edge of the disk, how fast was it spinning initially if it ends up moving at 7.45 rad/s after the bowl is removed?

Answer 1

2.969 rad/s

Step-by-step explanation:

Let's R be the radius of the spinning disk. By law of conservation of angular momentum we have before and after removing the bowl from disk

`A_1 = A_2`

`I_1\omega_1 = I_2\omega_2`

Where the mass moment of inertia for disk is
`MR^2/2` and for the bowl of fruit at edge of the disk is
`mR^2`. Assume the bowl is a point mass spinning at radius R

`I_1 = (MR^2)/(2) + mR^2 = 2.65R^2 + 4R^2 = 6.65R^2`

After removing the bowl, the moment of inertia is only the disk

`I_2 = (MR^2)/(2) = 2.65R^2`

Therefore we can substitute in for the equation of angular momentum conservation:

`6.65R^2\omega_1 = 2.65R^2*7.45`

Divide both sides by
`R^2` and we get

`6.65\omega_1 = 19.742`

`\omega_1 \approx 2.969 rad/s`