A solid cylinder rolls down an inclined plane of height `3m` and reaches the bottom of plane with angular velocity of `2sqrt2 rad//s`. The radius of cylinder must be [take `g=10m//s^(2)`]

Options

A. 5 cm
B. 0.5 cm
C. `sqrt(10)` cm
D. `sqrt(5)` m

Answer

D

Explanation

The relation between linear velocity `(v)` and angular velocity `(omega)` is
`v=r omegarArr r=(v)/(omega)`
Total kinetic energy `= (1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
`=(1)/(2)mv^(2)+(1)/(2)xx(1)/(2)mr^(2)omega^(2)=(1)/(2)mv^(2)+(1)/(4)mv^(2)=(3)/(4)mv^(2)`
`(3)/(4)mv^(2)=mg.3rArrv=2 sqrt(10)`
`because omegar=vrArrr=(v)/(omega)=(2sqrt(10))/(2 sqrt(2))=sqrt(5)` m.