How Does Angular Momentum Affect a Ball Rolling Up an Incline?

[myko]

A hollow sphere with mass $$M$$ , radius $$R$$ , and moment of inertia $$I=2/3MR^2$$ about its center rolls without slipping with a initial center of mass speed $$v$$ towards a fixed ramp. It then rolls without slipping up the ramp. The ramp forms an angle $$\theta$$ with the horizontal. The coefficient of static friction between the ramp and ball is$$\mu_s$$ , and the coefficient of kinetic friction is $$\mu_k$$ .
How long will it take for the ball to reach its maximum height?

I can find the height at which the ball stops using energy equation. But not really sure how to calculate time it takes to get there. I always get that $$t=v/g \sin\theta$$, but I think it is wrong

 

[olivermsun]

One thing you can do is write out the energy equation and notice that you have a relationship between velocity and height, from which you can get velocity or height vs time.

 

[myko]

Ok. So here is the energy equation: $$1/2Mv^2+1/3Mv^2=Mgh$$
from which I get this relationship; $$h=\frac{5}{6g}v^2$$
And now, I supose, I have to express the distance traveled on the incline.$$d=h/\sin\theta$$
Now I could find the acceleration $$a=v^2/2d=v^2sin\theta/2h$$
and from here I can get the time that it takes to slow down $$t=v/a=v\frac{2h}{v^2sin\theta}$$
is this correct?

 

[Nathanael]

Yes that is correct $t=\frac{5v}{3gsin(\theta )}$

I did it a much different way, though. (Involving the frictional force.)

Your way was easier, but you assumed constant acceleration.Edit:
If you assume constant acceleration, then the time is the distance (d) divided by the average velocity (in this case v/2)

So you get $t=\frac{2d}{v}=\frac{2h}{vsin(\theta )}=\frac{5v}{3gsin(\theta )}$

 

[haruspex]

Taken literally, there is a sudden change from a level flat surface to a slope. If so, there's a sudden impulse when the ball hits the slope. Work will not be conserved. You can use a different conservation law to figure out what happens in that instant.

 

[myko]

Yes that is correct $t=\frac{5v}{3gsin(\theta )}$

I did it a much different way, though. (Involving the frictional force.)

Your way was easier, but you assumed constant acceleration.


Edit:
If you assume constant acceleration, then the time is the distance (d) divided by the average velocity (in this case v/2)

So you get $t=\frac{2d}{v}=\frac{2h}{vsin(\theta )}=\frac{5v}{3gsin(\theta )}$

Could you show how did you do it?

 

[Nathanael]

Could you show how did you do it?

Sure.



The net force on the object as it rolls up the hill is $mgsin(θ)$ (gravity) minus $F_f$ (friction)
(The reason the friction is in the opposite direction becomes clear if you imagine what the direction of the torque is)
So you have this equation for the acceleration:
$a=gsin(\theta )-\frac{F_f}{M}$

I will call this the "first equation"

...

We know that the angular acceleration is the linear acceleration of the CoM (what I've called "a") divided by the radius of the ball
(this is because it rolls without slipping)
$\alpha = \frac{a}{R}$ ... (Where $\alpha$ is the angular acceleration)
We know the frictional force is what is causing the torque.
$RF_f=\tau$
We also know that the torque is the angular acceleration multiplied by the rotational inertia.
$\tau =I{\alpha}$

So, this all leads to the following equation:
$RF_f=\frac{2}{3}MR^2\frac{a}{R}$
Which becomes:

$F_f=\frac{2}{3}Ma$



Plugging that into the "first equation" we finally get:
$a=gsin(\theta )-\frac{2}{3}a$

Solving for a we get:

$a=\frac{gsin(\theta )}{1+\frac{2}{3}}=\frac{3gsin(\theta )}{5}$

And of course, $\frac{v}{a}=t$ so we get

$t=\frac{5v}{3gsin(θ)}$

 

[Nathanael]

Now that seems like a difficult way to do the problem, but it took a lot more effort to explain it than to actually do it :tongue:

The other method ($t=\frac{distance}{avrg.vel}=\frac{2d}{v}$) is much faster, but the only thing is, (if you didn't have an answer sheet) you would have to justify your assumption that the acceleration is constant.

 

[olivermsun]

It's an energy equation with kinetic energy proportional to distance traveled. Would it be fair to infer constant acceleration from that alone?

 

[haruspex]

It's an energy equation with kinetic energy proportional to distance traveled. Would it be fair to infer constant acceleration from that alone?

Two things...
First, did you notice my post #5? I can't be sure the question intends you to take this into account, but I feel you should.
Secondly, once you get past that, you can use conservation of energy without having to worry about forces or constancy of acceleration. Just use that the loss in KE in coming to a stop will equal the gain in PE. (Nathanael, you're making it much harder than it needs to be.)

 

[olivermsun]

I seriously doubt the question is concerned with the sudden onset of the inclined plane. Otherwise maybe you'd also have to take into account the finite radius of the sphere and how that interacts with the "corner."

I was just talking about justifying the constant acceleration to get the travel time (in response to Nathanael's post).

 

[Nathanael]

Just use that the loss in KE in coming to a stop will equal the gain in PE. (Nathanael, you're making it much harder than it needs to be.)

Sorry, but how does that give you the time?

All of my posts had to do with the time aspect of the problem (because conservation of energy disregards time)


Edit:

It's an energy equation with kinetic energy proportional to distance traveled. Would it be fair to infer constant acceleration from that alone?

I think so.

Can I prove it? Nope.

 

[olivermsun]

I think so.

Can I prove it? Nope.

Ah.

I was thinking along these lines:
$$\begin{align}
E_\mathrm{total} &= RE + KE + mgh & \\
&= \dfrac{5}{3}KE + mgh \\
KE &= \dfrac{3}{5} E_\mathrm{total} - \dfrac{3}{5} mgh = \mathrm{const.} - \dfrac{3}{5} mgh.
\end{align}$$
This is just the usual KE equation for falling under constant acceleration, except here it's as if the gravitational force is $\dfrac{3}{5}g$.

To be more rigorous, you can take derivatives with respect to time on both sides to get
$$\begin{align}
v \dfrac{dv}{dt} &= -\dfrac{3}{5} g v\sin\theta \\
\dfrac{dv}{dt} &= -\dfrac{3}{5} g \sin\theta = \mathrm{const.}
\end{align}$$.

 

[haruspex]

Sorry, but how does that give you the time?

No I'm sorry, you're right - I misremembered what the question was asking for.
But maybe one could simply observe that the forces and torques are constant, therefore the acceleration is constant? Is that slightly easier than solving the force equation?

I seriously doubt the question is concerned with the sudden onset of the inclined plane.

I don't see why it's so unreasonable. You are given all the information to do it.

Otherwise maybe you'd also have to take into account the finite radius of the sphere and how that interacts with the "corner."

That's not an also - it's exactly what I'm referring to. It's not particularly hard, and the effect is quite significant.

 

[olivermsun]

Because it's an extra complication that doesn't really have anything to do with the main concept, which is
$$RE + KE + PE = \mathrm{const.}$$
You can always make the effect of the corner negligible by specifying that the radius is "small" or the sphere has always been on the incline, etc., i.e., the usual idealizations in this kind of problem.

 

[Nathanael]

But maybe one could simply observe that the forces and torques are constant, therefore the acceleration is constant? Is that slightly easier than solving the force equation?

Yes, I completely agree that the constant-acceleration-method is far far superior.

I did not mean to imply that it would be difficult to show that acceleration is constant, I just wanted to point out that the assumption was made, and so the OP should recognize that it's been made.(The only reason I made that long post involving force equations is because the OP requested my alternate solution.)

(And the only reason that I solved it that way is that I'm imperfect and I overlooked the simplest solution.)

 

[haruspex]

You can always make the effect of the corner negligible by specifying that the radius is "small"...

No, that doesn't fix it. The fraction of energy lost depends on the angle of the slope, not on the radius of the sphere.

 

[olivermsun]

It fixes the problem that the sphere encounters the corner before the center is above the incline.
Other than that, what is the problem you perceive with respect to the energy equation?

 

[myko]

Yes, I completely agree that the constant-acceleration-method is far far superior.

I did not mean to imply that it would be difficult to show that acceleration is constant, I just wanted to point out that the assumption was made, and so the OP should recognize that it's been made.(The only reason I made that long post involving force equations is because the OP requested my alternate solution.)

(And the only reason that I solved it that way is that I'm imperfect and I overlooked the simplest solution.)

Ty Nathanael for your solution. I ,realy think, it is also a good and easy way to solve this. Actualy I think this is the way the problem was intended to be solved in the test I took. Thankyou again. What regard the asumption of constant acceleration, constant force implies constant acceleration. So better call it "consequence" rather than asumption. :)

 

[haruspex]

It fixes the problem that the sphere encounters the corner before the center is above the incline.

Only if you assume the transition from horizontal to incline has some smoothness. If it is an abrupt change, it makes no difference how small the radius is.
For the present problem (hollow sphere), the fraction of energy lost looks like this:
30^o: 15%
45^o: 32%
60^o: 51%
90^o: 84%

So in general it is an important consideration.
Now, it's quite possible that the problem setter also overlooked its significance. If so, I regard it as a poorly specified question.

 

[olivermsun]

I guess I don't fully understand your point. I thought you were pointing out that the sphere actually starts rolling uphill slightly before the center is over the beginning of the incline. You seem to be concerned about the impulse as the kinetic energy of the ball which was formerly rolling horizontally is suddenly redirected into slightly uphill kinetic energy. Why can't this be assumed to be an "elastic but still sticking to the surface" interaction for the purposes of the problem?

 

[haruspex]

Why can't this be assumed to be an "elastic but still sticking to the surface" interaction for the purposes of the problem?

Because that doesn't help. It would mean the lost energy is now in the form of oscillations in the ball, to be released as soon as the magic making it stick to the surface goes away.
I'm not being facetious. All you have to do is consider conservation of angular momentum about the point of impact. This will prove there's a loss of kinetic energy, and no tricks like the one you propose will overcome it.

 

[olivermsun]

I don't think that the student is usually required to consider the oscillations in the ball, or the ramp/bumper material, or anything else like that. As you well know, these questions are posed in "ideal model-world." There are any number of these problems where the behavior is unrealistic and cannot be saved by any "tricks," although a well-used real-world-physics defense is that we never actually have truly instantaneous collisions or transitions in the real world.

But perhaps you will indulge my lack of understanding. Would you please write out what you mean, in equations?

 

[rcgldr]

What was the point of stating both static and kinetic cofficients of friction since the problem states that the ball rolls up the incline without slipping?

As for the collision with the incline, the situation could be idealized to a totally elastic collision that somehow doeen't involve bouncing or the ball not internally twisting creating an angular oscillation that never dissipates or just happens to transition into zero internal torque at the moment the ball stops rolling.

 

[haruspex]

What was the point of stating both static and kinetic cofficients of friction since the problem states that the ball rolls up the incline without slipping?

To give the student more opportunities to go wrong. I approve of this style of question. In the real world, much irrelevant data is available.

As for the collision with the incline, the situation could be idealized to a totally elastic collision that somehow doeen't involve bouncing or the ball not internally twisting creating an angular oscillation that never dissipates or just happens to transition into zero internal torque at the moment the ball stops rolling.

How, other than supposing a smooth transition in the surface shape (which ought to be stated, not presumed)? Bear in mind that such an idealisation for an abrupt change in slope also means violating a conservation law. How would you justify that?

 

[rcgldr]

Bear in mind that such an idealisation for an abrupt change in slope also means violating a conservation law. How would you justify that?

It really can't be done, an idealized 100% elastic ball would end up bouncing up the ramp due to the collision resulting in a bouncing motion (at least at the center of mass in the case of a very soft but elastic ball), and that bouncing motion would consume some of the energy.

What happens during the collision is that momentum of the ball, ramp, and whatever the ramp is attached to, such as the earth, is conserved. Any loss of momentum of the ball is gained by the earth, but since the Earth is massive, the change in velocity of the Earth is tiny, and almost all of the energy would go into the ball. Still, if the collision is 100% elastic, I don't see a way to avoid bouncing being an issue.

 

[olivermsun]

I called it "elastic but still sticking" because I think the problem is meant in some idealization where 1) energy is conserved but 2) the ball stays attached to the slope without slipping, bouncing, flying off into space, etc.

Are normal "elastic" collision idealizations really "justified"? It seems to me that if a pool ball bounces perfectly elastically off a rubber bumper, then surely some energy is lost to the bumper, which continues to oscillate after the pool ball leaves, which is a contradiction…

We commonly prescribe a idealized behavior like this with movement of "particles" over arbitrarily-shaped bottoms. In the usual formulation it doesn't even conserve mass, strictly speaking. Surely this cannot be justifiable.

 

[haruspex]

It really can't be done, an idealized 100% elastic ball would end up bouncing up the ramp due to the collision resulting in a bouncing motion (at least at the center of mass in the case of a very soft but elastic ball), and that bouncing motion would consume some of the energy.

What happens during the collision is that momentum of the ball, ramp, and whatever the ramp is attached to, such as the earth, is conserved. Any loss of momentum of the ball is gained by the earth, but since the Earth is massive, the change in velocity of the Earth is tiny, and almost all of the energy would go into the ball. Still, if the collision is 100% elastic, I don't see a way to avoid bouncing being an issue.

I'm confused - I thought you previously argued that it could retain all its KE in some way. Now you seem to be agreeing it can't.

Btw, I was not referring to conservation of momentum. Use conservation of angular momentum about the point where the ball makes contact with the ramp. The impact does not affect that. By assuming the ball remains in rolling contact with the ramp (i.e. 100% inelastic, and sufficient coefficient of static friction), you can deduce the new speed.

 

[haruspex]

I called it "elastic but still sticking" because I think the problem is meant in some idealization where 1) energy is conserved but 2) the ball stays attached to the slope without slipping, bouncing, flying off into space, etc.

That is your interpretation of the problem, and it may be that the problem setter had that in mind. But not bouncing, not slipping, not losing any KE implies conservation of angular momentum is significantly violated. That's not just an idealisation - it's an error.

Are normal "elastic" collision idealizations really "justified"? It seems to me that if a pool ball bounces perfectly elastically off a rubber bumper, then surely some energy is lost to the bumper, which continues to oscillate after the pool ball leaves, which is a contradiction…

Of course no macroscopic collisions in the real world are perfectly elastic. That is an idealisation, but you don't have to violate any other law in presuming it.

 

[ehild]

All you have to do is consider conservation of angular momentum about the point of impact. This will prove there's a loss of kinetic energy, and no tricks like the one you propose will overcome it.

Can you explain why does the angular momentum abut the point of contact conserved when the ball starts to roll upward? Which point do you speak about? The one just the impact? About that, the impact force provides torque. At the instant of the impact the ball loses contact with the horizontal ground so it gets subjected by the torque of its weight.

ehild