Calculating the Speed of a Cylinder on a Frictional Ramp

[delecticious]

Homework Statement

A uniform solid cylinder (m=0.230 kg, of small radius) is at the top of a similar ramp, which has friction. The cylinder starts from rest and rolls down the ramp without sliding and goes around the loop. Find the speed of the cylinder at the top of the loop.

Homework Equations

PE - potential energy
KE - kinetic energy
KEr - rotational kinetic energy
I - moment of inertia
w - omega

KE0 + PE0 + KEr0 = KEf + PEf + KErf

The Attempt at a Solution

At first I was tripped up over the small radius part, until I realized that the radius would cancel out with final height of the potential energy and I deduced that the initial rotational and kinetic energies would cancel out to get something to this fashion:

mgh0 = 1/2mv^2 + mg2R + 1/2Iw^2
mgh0 = 1/2mv^2 + mg2R + 1/2I(v^2/r^2)
mgh0 = 1/2mv^2 + mg + 1/2Iv^2

I plugged and chugged my numbers to get a velocity of 7.18, but then I realized I forgot to incorporate the friction force, so how would that fit into the mathematical equation?

 

[Doc Al]

mgh0 = 1/2mv^2 + mg2R + 1/2Iw^2
mgh0 = 1/2mv^2 + mg2R + 1/2I(v^2/r^2)

OK.

mgh0 = 1/2mv^2 + mg + 1/2Iv^2

Not sure what you did here. What happened to R and r?

I plugged and chugged my numbers to get a velocity of 7.18, but then I realized I forgot to incorporate the friction force, so how would that fit into the mathematical equation?

I didn't do the calculation myself, but you need to double check your formulas. What did you use for I?

Since the cylinder rolls without slipping, no energy is lost to friction: Mechanical energy is conserved.

 

[delecticious]

OK.

Not sure what you did here. What happened to R and r?

sorry, R=r so I should have had them in the same caps. Since they're the same I canceled them out

I didn't do the calculation myself, but you need to double check your formulas. What did you use for I?

I = 1/2MR^2 <---- I looked it up in my book and it said the moment of inertia for solid cylinders is that

Since the cylinder rolls without slipping, no energy is lost to friction: Mechanical energy is conserved.

so does that mean the fact that there is friction is negligible in this situation?

 

[Doc Al]

sorry, R=r so I should have had them in the same caps. Since they're the same I canceled them out

R is the radius of the loop; r is the radius of the rolling cylinder. They are not the same. (And even if they were, how would they cancel?)

Rewrite the final equation that you used to get your answer.

I = 1/2MR^2 <---- I looked it up in my book and it said the moment of inertia for solid cylinders is that

Good.

so does that mean the fact that there is friction is negligible in this situation?

It's not that the friction is negligibly small, and thus can be ignored. Friction is essential--it's what makes the cylinder roll instead of slide. But from an energy point of view, the friction does no work so energy is still conserved. (So it has no effect on your energy equation.)

 

[delecticious]

R is the radius of the loop; r is the radius of the rolling cylinder. They are not the same. (And even if they were, how would they cancel?)

looks like goofed here. Ok so R stays, but r isn't given so somehow it drops out, but I'm not sure how it would.

 

[Doc Al]

Small r appears in the formula for I.

 

[delecticious]

oh, no duh, thanks for the help