# Ring and solid sphere rolling down an incline - rotation problem

Gold Member

## Homework Statement

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?

## Homework Equations

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?

haruspex
Homework Helper
Gold Member
minimum coefficient of friction required for hollow sphere to roll is 2tanθ/5
Is that right? What value of n are you plugging in?
Friction is sufficient to prevent slipping of hollow sphere-
Should this say "just sufficient"?

Gold Member
Is that right? What value of n are you plugging in?
n is 2/3 for a hollow sphere. (2/3)/(1+(2/3)) gives 2/5 so 2/5tanθ.
Should this say "just sufficient"?
It doesn't say that. But knowing the way around this book, I can safely say that was most probably the meaning. Otherwise there'd be two cases- one where all are rolling and one where only the ring isn't; in any case the solid sphere seems to win both races

haruspex
Homework Helper
Gold Member
n is 2/3 for a hollow sphere. (2/3)/(1+(2/3)) gives 2/5 so 2/5tanθ.
Sorry, I misread it as solid sphere.

Gold Member
Sorry, I misread it as solid sphere.
so everything's correct?

haruspex
Homework Helper
Gold Member
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?
Yes, this all looks right, but I think you can make it a bit simpler by comparing accelerations instead of times: (5/7)g sin(θ) for the solid sphere, (3/5)g sin(θ) for the ring.

Gold Member
Yes, this all looks right, but I think you can make it a bit simpler by comparing accelerations instead of times: (5/7)g sin(θ) for the solid sphere, (3/5)g sin(θ) for the ring.
Alright thank you

haruspex
Homework Helper
Gold Member
Should this say "just sufficient"?
On reflection, it does not need to. Increasing it will slow the ring further but not affect the others.

• jbriggs444
Gold Member
On reflection, it does not need to. Increasing it will slow the ring further but not affect the others.
ah! right.