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. 1370 J
B. 1630 J
C. 1470 J
D. 1570 J

Answer

C

Explanation

`n_(1)=(600)/(60)="10 r.p.s, "n_(2)=(300)/(60)="5 r.p.s"`
`K=(1)/(2)Iomega^(2)=(1)/(2)I(2pin)^(2)=2pi^(2)n^(2)I`
`thereforeK=2pi^(2)n^(2)((9.8)/(pi^(2)))="19.6n^(2)`
`because n_(1) gt n_(2) " "therefore " "DeltaK=K_(1)-K_(2)=19.6[n_(1)^(2)-n_(2)^(2)]`
`=19.6(100-25)`
`=75xx19.6=1470J`
Work done = decrease in K.E. = 1470J