A thin metal disc of radius 0.25 m and mass 2 kg starts from rest and rolls down an inclined plane. If its rotational kinetic energy is 4 J at the foot of the inclined plane, then its linear velocity at the same point is

Options

A. `1.2 ms^(-1)`
B. `2sqrt2 ms^(-1)`
C. `20 ms^(-1)`
D. `2 ms^(-1)`

Answer

(b)

Explanation

`r=0.25m`,
`m=2kg` (Given)
Rotational Kinetic energy,
`K.E=4`
`(1)/(2) Iomega^(2)=4`
implies `(1)/(2)xx(1)/(2)mr^(2)xx((v)/(r))^(2)=4`
`(1)/(2)xx(1)/(2)xx2xx0.25^(2)xx(v^(2))/(0.25^(2))=4`
implies `v^(2)=8`
`v=2sqrt2m//s`.