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. `1467` J
B. `1452` J
C. `1567` J
D. `1470` J

Answer

D

Explanation

`I=(9.8)/(pi^(2))"kgm"^(2), f=(600)/(60)=10` rps
`omega_(0)=2pixx10=20pi` rad/s
`omega=2pixx(300)/(60)=10 pi` rad/sec
Work done = Change is KE
= `(1)/(2)I(omega_(0)^(2)-omega^(2))`
= `(1)/(2)xx(9.8)/(pi^(2))xx4pi^(2)[10^(2)-5^(2)]`
= `1470` J .