A solid sphere is projected up along an inclined plank of inclination `theta=30^@` with a speed `v=2ms^(-1)`. If it rolls without slipping find the maximum distance traversed by it (`g=10 ms^(-2))`

Explanation

Let the solid sphere of mass .m. and radius r traverse a distance l along the inclined plane. Since it rolls without sliding its initial K.E. is given as
`KE_(i) = (1)/(2) mv^(2)(1+(k^(2))/(r^(2))),"where k" = sqrt((2)/(5))r`
`= (1)/(2)mv^(2) (1+(2)/(5)) = (7//10)mv^(2)`

As it comes to rest attaining a height h = l sin `theta`, its final KE = 0
`Delta KE = -(7)/(10) mv^(2)" "...(a)`
Change in gravitational, `P.E. = mgh" "...(b)`
Conservation of energy yields,
`(Delta PE) + (Delta KE) = 0`
`rArr" "mgh - (7//10) mv^(2) = 0`
`rArr" "h = (7v^(2))/(10g) = (7 xx (2)^(2))/(10 xx 10) = 0.28 m`
`therefore" "l = "h cosec" theta = "0.28 cosec 30"^(@) = 0.56 m`