A ring of mass 0.3 kg and radius 0.1 m and a solid cylinder of mass 0.4 kg and of the same radius are given the same kinetic energy and released simultaneously on a flat horizontal towards a wall which is at the same distance from the ring and the cylinder. The rolling friction in both cases is negligible. The cylinder will reach the wall first.

Explanation

Total energy of the ring
`=(K.E.)_(R_otation) + (K.E.)_(T_ranslational)`
`=(1)/(2) Iomega^2 + (1)/(2) mv_c^2`
`=(1)/(2)xxmr^2omega^2+(1)/(2)m(romega)^2 (:. I =mr^2, v_c =romega)`
`=mr^2omega^2`
Total kinetic energy of the cylinder
`=(K.E.)_(R_otation)+(K.E)_(T_ranslational)`
` = (1)/(2) i' omega^2 + (1)/(2)Mv_c^2`
`=(1)/(2)((1)/(2) Mr^2))omega^2 +(1)/(2)M(romega)^2`
`=(3)/(4)Mr^2omega^2` .....(i)
Equating (i) and (ii)
`mr^2omega^2 = (3)/(4)Mr^2 omega^2`
`rArr (omega^2)/(omega^2) = (4m)/(3M)=(4)/(3)xx(0.3)/(0.4) =1`
`rArr omega = omega`
Both will reach at the same time.