Question

A solid disk is rolling without slipping on a level surface at a constant speed of `2.00m//s`. How far can it roll up a `30^(@)` ramp before it stops? (take `g=9.8m//s^(2))`

Answer 1

The moment of inertia of the sphere about its axis of rotation is `I = (2)/(5) mr^(2)`. Suppose it rises to height `h` before stopping. Using the conservation of energy, we get

`mgh= (1)/(2)mv^(2)+(1)/(2)Iomega^(2)=(1)/(2)mv^(2)+(1)/(2)((2)/(5)mr^(2))omega^(2)`
`=(1)/(2)mv^(2)+(1)/(5)mv^(2)=(7)/(10)mv^(2) or h=(7v^(2))/(10g)`
`:.` The distance rolled up on the ramp is
`s=(h)/(sin theta)=(7v^(2))/(10g sin theta)=(7("2 ms"^(-1))^(2))/(10(9.8 "ms"^(-2))xx1//2)=(4)/(7) "metre" ~= 57 "cm"`