A thin hollow cylinder, open at both ends and weighing `5kg` (a) slides with a speed of `10 m//s` without rotating (b) rolls with the same speed without slipping. Compare the kinetic energies of the cylinder in the two cases.

Explanation

Here, `m = 5 kg, v = 10 m//s`
(a) When cylinder slides without rotating, its energy is wholly translatory.
`E_(1) = (1)/(2) m upsilon^(2) = (1)/(2) xx 5 xx 10^(2) = 250 J`
(b) When cylinder rolls without slipping, it has both translational and rotational `K.E`.
`E_(2) = (1)/(2) m upsilon^(2) + (1)/(2) I omega^(2)`
But `I = m r^(2) and r omega = upsilon`
`:. E_(2) = (1)/(2)m upsilon^(2) + (1)/(2)(m r^(2)) omega^(2) = (1)/(2) m upsilon^(2) + (1)/(2) m upsilon^(2) = m upsilon^(2) = 5 xx 10^(2) = 500 J :. (E_(1))/(E_(2)) = (250)/(500) = (1)/(2)`