A solid sphere, hollow sphere, disk and ring are released simultaneously from top of a incline. Friction is sufficient to prevent slipping of hollow sphere- what will reach the bottom first?
a in pure rolling down an incline=gsinθ/(1 + I/mR^2)
The Attempt at a Solution
So first I determined which objects are pure-rolling and which are slipping. Lets say I/mR^2=n which is just the coefficient with the mr^2 term for convenience. Now from the above formula acceleration for a body in pure rolling is given by gsinθ/(1+n). Now, mgsinθ-frictionforce=mgsinθ/(1+n) therefore frictional force=nmgsinθ/(1+n). Using frictional force<=μN and N=mgcosθ we get μ>=ntanθ/1+n. So minimum coefficient of friction required for hollow sphere to roll is 2tanθ/5, plugging n for the other bodies we see that all the bodies except the ring will be pure rolling. Now when all 4 are pure rolling, solid sphere has highest value of acceleration so reaches first, but now ring is slipping so the competition is between the solid sphere and the ring.
Using s=(1/2)at^2 (s is same for both bodies) t for solid sphere is sqrt(14s/5gsinθ). I'm getting very confused finding the t for the ring though, any help would be appreciated-thank you
Edit: So, the acceleration for the ring is g(sinθ-μcosθ), plugging the minimum value of μ we found for hollow sphere, we get t=sqrt(10s/3gsinθ) which indicates the solid sphere will reach first. Is that correct?