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)`
`therefore k^(2)=(2)/(5)R^(2)`
Hence, `v = sqrt((2gh)/((1+(2)/(5))))`
`= sqrt((2gh xx 5)/(7))`
`= sqrt((2xx10xx20xx5)/(7))`
`= sqrt(285.714)`
`= 16.90 m//s`.