A disc of moment of inertia `9.8//pi^(2)kg m^(2)` is rotating at 600 rpm. If the frequency of rotation changes from 600 rpm to 300 rpm, then what is the work done ?

Options

A. 1567 J
B. 1452 J
C. 1467 J
D. 1632 J

Answer

C

Explanation

`omega = KE_(2)-KE_(1)=(1)/(2)I (omega_(2)^(2)-omega_(1)^(2))`
`= (1)/(2)xx(9.8)/(pi^(2))xx4 pi^(2)(n_(2)^(2)-n_(1)^(2))`
`= 4.9xx4(25-100)`
`=4.9xx4xx75`
`= 4.9xx300`
`= 1470.0`
`= 1467 J`