A solid cylinder at rest at top of an inclined plane of height `2.7 m` rolls down without slipping. If the same cylinder has to slide down a frictionaly inclined plane and acquires the same velocity as that acquired by centre of mass of rolling cylinder at the bottom of the incline, what should be the height of inclined plane ?

Explanation

Here, `h_(1) = 2.7m, h_(2) = ?`
When the solid cylinder rolls without slipping, its total kinetic energy
`K = K_(t) + k_(r )`
`= (1)/(2)m upsilon^(2) + (1)/(2)I omega^(2)`
`= (1)/(2)m upsilon^(2) + (1)/(2)((1)/(2)m r^(2)) omega^(2)`
`K = (1)/(4) m upsilon^(2) + (1)/(4)m upsilon^(2) = (3)/(4)m upsilon^(2)`
But `K = mg h_(1) = (3)/(4)m upsilon^(2)` ..(i)
In sliding down a height `h_(2)`.
`mg h_(2) = (1)/(2)m upsilon^(2)` ...(ii)
Dividing (ii) by (i), we get
`(h_(2))/(h_(1)) = (1//2)/(3//4) = (2)/(3)`
`h_(2) = (2)/(3)h_(1) = (2)/(3) xx 2.7 m = 1.8 m`