Question

A disk of mass 1.5 kg and radius 65 cm with a small mass of 0.04 kg attached at the edge is rotating at 1.9 rev/s. The small mass suddenly flies off of the disk. What is the disk's final rotation rate (in rev/s)? It says 2.00 is worng.

Answer 1

Using Angular momentum conservation:

Li = Lf

Ii*wi = If*wf

Ii = Initial moment of inertia of disk + small mass = I0 + Im

I0 = moment of inertia of disk = (1/2)*M*R^2 = (1/2)*1.5*0.65^2 = 0.316875 kg.m^2

Im = moment of inertia of small mass about rotation axis = m*R^2 = 0.04*0.65^2 = 0.0169 kg.m^2

I1 = 0.316875 + 0.0169 = 0.333775 kg.m^2

wi = initial angular velocity = 1.9 rev/sec

If = final moment of inertia of only disk = I0 = 0.316875 kg.m^2

So,

wf = final angular velocity = wi*(Ii/If)

wf = 1.9*(0.333775/0.316875)

wf = 2.00133 rev/sec = 2.0 rev/sec

Answer will be 2.0 rev/sec, may be you need final answer in two significant figures, So use 2.0 rev/sec, instead of 2.00 rev/sec