A body having a moment of inertia about its axis of rotation equal to 3 kg `m^(-2)` is rotating with angular velocity of 3 rad `s^(-1)`. Kinetic energy of this rotating body is same as that of a body of mass 27 kg moving with velocity v. The value of v is

Options

A. `2 ms^(-1)`
B. `1 ms^(-1)`
C. `1.5 ms^(-1)`
D. `0.5 ms^(-1)`

Answer

B

Explanation

Here, Moment of inertia of the body about its axis of rotation, `I=3 "kg m^(2)`
Angular velocity of rotation, `omega= 3 "rad" s^(-1)`
Kinetic energy of rotation of the body is
`K_(R)= 1/2 I omega^(2)`
Kinetic energy of body of mass `m(=27 "kg")` moving with velocity v is
`K_(T)= 1/2 mv^(2)`
As per question, `K_(R)=K_(T)`
`:.1/2I omega^(2)= 1/2 mv^(2)`
`1/2xx3xx3^(2)=1/2xx27xxv^(2)`
or `v^(2)=1` or `v=1 m s^(-1)`