Ring and solid sphere rolling down an incline - rotation problem

[Krushnaraj Pandya]

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. Let's 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]

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"?

 

[Krushnaraj Pandya]

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]

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.

 

[Krushnaraj Pandya]

Sorry, I misread it as solid sphere.

so everything's correct?

 

[haruspex]

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.

 

[Krushnaraj Pandya]

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]

Should this say "just sufficient"?

On reflection, it does not need to. Increasing it will slow the ring further but not affect the others.

 

[Krushnaraj Pandya]

On reflection, it does not need to. Increasing it will slow the ring further but not affect the others.

ah! right.