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. 1470 J
B. 1452 J
C. 1567 J
D. 1632 J

Answer

A

Explanation

According to work-energy theorem,
Work done = Change in rotational kinetic energy
`W=(DeltaKE_(f))_(1)-(DeltaKE_(r))_(2)" …(i)"`
But rotational kinetic energy, `K=(1)/(2)l omega^(2)`
From Eq. (i), we get
`W=(1)/(2)l omega_(1)^(2)-(1)/(2)l omega_(2)^(2)`
`=(1)/(2)l(omega_(1)^(2)-omega_(2)^(2))`
`"As, "omega=2pin`
Hence, we get `" "W=(1)/(2)l[(2pin_(1))^(2)-(2pin_(2))^(2)]`
`=(1)/(2)lxx4pi^(2)(n_(1)^(2)-n_(2)^(2))" ...(ii)"`
Given,`" "l=(9.9)/(pi^(2))kg-m^(2)`
`n_(1)="600 rpm = 10 rps"`
`n_(2)="300 rps = 5 rps"`
From Eq. (ii), we get
`W=(1)/(2)xx(9.8)/(pi^(2))xx4pi^(2)(10^(2)-5^(2))=1470J`