A solid sphere is rolling down an inclined plane without slipping of height 20 m. Calculate the maximum velocity with which it will reach the bottom of the plane `(g = 10 m//s^(2))`

Explanation

When the sphere rolls down, its potential energy changes to `K.E.` of rotation. Therefore,
`K.E.=P.E.`
`(1)/(2)mv^(2)(1+(k^(2))/(k^(2)))=mgh`
`v=sqrt((2gh)/((1+(k^(2))/(k^(2)))))`
Moment of inertia of solid sphere `=(2)/(5)MR^(2)`
`:. k^(2)=(2)/(5)R^(2)`
Hence, `v=sqrt((2gh)/((1+(2)/(5))))=sqrt((2ghxx5)/(7))=sqrt((2xx10xx20xx5)/(7))=sqrt(285.714)=16.90m//s`