A ring of diameter `0.4 m` and of mass `10 kg` is rotating about its axis at the rate of 2100 rpm. Calculate moment of inertia, angular momentum and rotational `KE` of the ring.

Explanation

Here, `r =(0.4)/(2) = 0.2 m, m = 10 kg`,
`n = (2100)/(60) rps = 35 rps`
`I = mr^(2) = 10(0.2)^(2) = 0.4 kg m^(2)`
`L = I omega = I (2pi n) = 0.4 (2 xx 3.14 xx 35)`
`= 88 kg m^(2) s^(-1)`
Rotational `KE = (1)/(2)I omega^(2) = (1)/(2) xx 0.4 (2pi xx 35)^(2)`
`= 9680 J`