Question

A disc of mass 2 kg and radius 0.2 m is rotating with angular velocity `30 "rad s"^(-1)`. What is angular velocity, if a mass of 0.25 kg is put on periphery of the disc ?
A. 24 rad `s^(-1)`
B. 36 rad `s^(-1)`
C. 15 rad `s^(-1)`
D. 26 rad `s^(-1)`

Answer 1

Correct Answer - A
If no external torque acts on a system of particles, then angular momentum of the system remains constant, i.e.,
`tau=0`
`rArr (dL)/(dt) = 0 rArr L = I omega =` constant
`r rArrI_(1)omega_(1)=I_(2)omega_(2)rArr(1)/(2)Mr^(2)omega_(1)=(1)/(2)(M+2m)r^(2)omega_(2)`....(i)
Here, `M=2kg,m=0.25kg,r=0.2m,omega_(1)=30 "rad s"^(-1)`
Hence, we get after putting the given values in Eq. (i)
`(1)/(2)xx2xx(0.2)^(2)xx30=(1)/(2)xx(2+2xx0.25)(0.2)^(2)xxomega_(2)`
or `1.2=0.05omega_(2)or omega_(2)=24 "rad s"^(-1)`.