A hollow cylinder rolls (about its geometrical axis) without slipping on an inclined plane of angle `theta`. Find its linear acceleration in the direction parallel to the surface of the inclined plane.

Explanation

The potential energy of a hollow cylinder at height `h=mgh`
When it rolls on slope its potential energy converted into kinetic energy.
`therefore` From the law of conservation of mechanical energy.
`mgh=(1)/(2)Iomega^(2)+(1)/(2)mv^(2)`
`therefore mgh=(1)/(2)mr^(2)omega^(2)+(1)/(2)mv^(2)" "[because I=mr^(2)]`
`therefore gh=(1)/(2)(v)^(2)+(1)/(2)v^(2)`
`therefore gh=v^(2)....(1)`
Now eqn. of linear motion,
`2ad=v^(2)-v_(0)^(2)`, Taking `v_(0)=0`
`2ad=v^(2)`
`therefore 2ad=gh` [`because` from eqn. (1)]
`therefore a=(gh)/(2d)`
Now from fig. `(h)/(d)=sintheta`
`therefore a=0.5g sintheta`