(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two cylinders (radius R) start from rest and roll without slipping down ramps. They both have the same masses and moments of inertia. The ramps are both at the same angle. One cylinder (A) is directly on the ramp. The other (B) is hanging from two smaller cylindrical bits (radius r) which stick out from either side and are in contact with the ramp. Which reaches the bottom first?

2. Relevant equations

τ = I α

τ = F r

mgh

1/2 mv^2

1/2 Iω^2

3. The attempt at a solution

I used torque to figure out which would have the greater ω. The force providing torque is friction, which must be the same for each cylinder (same mass, same normal force). The difference is in r; so the cylinder with the smaller r-value must have the smaller torque and thus the smaller ω at any moment in time. I.e.:

τ = F_f r = I α = I Δω/Δt

ω(t) = (F_f t / I) r

So it would seem that cylinder B, with the smaller r, must also have a smaller ω.

I know that the following is true for both cylinders:

mgh = 1/2 mv^2 + 1/2 Iω^2

So it would seem that cylinder B must have a greater v, because it has a smaller ω.

But I'm confused for a few reasons:

1. For rolling without slipping, v_cm = r ω. So if ω is smaller, so is v. This (obviously) contradicts the conclusion that smaller ω means greater v.

2. If friction is providing torque, it is increasing the rotational kinetic energy of the cylinders. Does that then mean that it is doing work? That would make the energy equation above incorrect (since it assumes zero work is done).

3. When I imagine the problem in my head, cylinder B seems like it should spin faster than cylinder A.

Any insight into this question is very much appreciated!

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# Two cylinders rolling without slipping

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