# Moving down a ramp

JessicaHelena

## Homework Statement

Please look at the problem attached as a screenshot.

## Homework Equations

Assuming frictionless, Ei = Ef, which means objects that are the same will end up in the same heights (so we can group A&C, B&D, and E&F).
For A&C and E&F, mgh = KE_rot + KE_trans
For B&D, it is mgh = KE_trans.
Also, v = rw (I know it's omega, but for convenience, I'll write it as w).
I _solid sphere = 2/5mr^2
I_hollow sphere = 2/3mr^2

## The Attempt at a Solution

After all these equations set up, now it's pretty much plugging things in. For A&C ,

mgh = 1/2mv^2 + 1/2Iw^2
mgh = 1/2mv^2 + 1/5mr^2w^2
gh = 1/2v^2 + 1/5 r^2(v/r)^2
gh = 1/2v^2 + 1/5v^2 = 7/10v^2
so v^2 = 10gh/7
then KE at the end is then 1/2mv^2 = 5mgh/7, and that can be converted to the new GPE. so the height will be 5/7 the original height.

For E&F, I can use a similar process, only now I = 2/3mr^2. That gives me a KE of 3/5mgh, so the new (final) height will be 3/5 the original height.

For the blocks (B&D), there's only KE_trans, so Ei = Ef and mgh = 1/2mv^2. With no rot KE to lose trans KE to, the final height should be the same as the original height.

Thus, in order, it is B&D, A&C, E&F.

Could someone please check if I don't have any flaws in my reasoning?

#### Attachments

• Screen Shot 2018-11-17 at 6.36.50 PM.png
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Where does the rotational energy go for the sphere on the upward slope?

JessicaHelena
Oh... so would it be that all objects (regardless of their shape) would end up with a same height that's identical to their original heights?

CWatters
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Oh... so would it be that all objects (regardless of their shape) would end up with a same height that's identical to their original heights?

Yes, but can you explain what happens on the up-slope? You need a free-body diagram to show what gravity and friction are doing.

JessicaHelena
It's accelerating down the slope even though it will continue moving up. So would that mean the objects don't quite reach the same height?

Using F_net = ma, I get that a = g*sin(theta), and I guess we could use v^2 = v_0^2 + 2ax (Where v = v^2 = 0) to get the x and then use trig to get the actual height, but wouldn't there be a better way using energy?

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It's accelerating down the slope even though it will continue moving up. So would that mean the objects don't quite reach the same height?

Using F_net = ma, I get that a = g*sin(theta), and I guess we could use v^2 = v_0^2 + 2ax (Where v = v^2 = 0) to get the x and then use trig to get the actual height, but wouldn't there be a better way using energy?

You can use energy considerations as an explanation (and not talk about forces). But, it's interesting to consider the forces as well:

On the downslope a sphere accelerates more slowly if it rolls (instead of slips).

On the upslope, the sphere must decelerate more slowly than it would under gravity alone in order to reach the same height. How does this happen?

JessicaHelena
Wouldn't it be similar to the downslope scenario — because the sphere is still rolling uphill, there'd be less of translation velocity, so it would decelerate more slowly as well?

But I'm a little lost, to be honest, when you say "On the downslope a sphere accelerates more slowly if it rolls (instead of slips)." It would certainly have less v_trans, but I don't think accelerating more quickly/slowly has much to do with the magnitude of v, and in that case, my answer above would be wrong too.

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Wouldn't it be similar to the downslope scenario — because the sphere is still rolling uphill, there'd be less of translation velocity, so it would decelerate more slowly as well?

But I'm a little lost, to be honest, when you say "On the downslope a sphere accelerates more slowly if it rolls (instead of slips)." It would certainly have less v_trans, but I don't think accelerating more quickly/slowly has much to do with the magnitude of v, and in that case, my answer above would be wrong too.

You started off with an answer that said: B&D, then A&C, then E&F.

Let's start again. Why don't you give your revised answer and why.