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 surface 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 `=(KE)_("Rotation")+(KE)_("Translational")`
`=1/2Iomega^(2)+1/2mV_(c)^(2)`
`:.I=mr^(2), V_(c)=romega`
Total `KE` of the cylinder
`=(KE)_("Rotational")+(KE)_("Translational")`
`=1/2=I^(')omega^('2)+1/2MV_(c)^(2)`
`=1/2(1/2Mr^(2))omega^('2)+1/2M(romega^(')^(2)`
`=3/4Mr^(2)omega^('2)` .........i.
Equation Eqs. i and ii `mr^(2)omega^(2)=3/4Mr^(2)omega^('2)`
`implies(omega^('2))/(omega^(2))=(4m)/(3M)=4/3xx0.3/0.4=1`
`=omega^(')=omega`
Both will reach at the same time.