Question

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))`

Answer 1

When the sphere rolls down, its potential energy changes to KE of rotation. Therefore, `KE=PE`

`(1)/(2)mv^(2)(1+(K^(2))/(R^(2)))=mgh`
`rArrv=sqrt((2gh)/((1+K^(2)//R^(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)=16.90 m//s`.