Sphere sliding up a step - Inelastic Collision

[Starkrod]

Homework Statement

A sphere of radius R is rolling without slipping with a velocity v and collides inelastically with a step of height h < R. ¿What is the minimum velocity for which the sphere will be over the step?

Homework Equations

Total kinetic energy (maybe):
\frac{1}{2}I\omega^{2}+\frac{1}{2}mv^2
Gravitational potential energy:
mgh
Moment of inertia about a point a distance d from the CM from de center of mass:
I=I_{cm}+md^{2}
Rolling without slliping
v=\omega R
3. A discussion of the problem
This problem is confusing to me: I know the collision is inelastic so there is no conservation of kinetic energy, there is torque when the sphere pivots the step so the angular momentum is not conserved either, the linear momentum doesn't give a lot information and even though I tried to see what condition needs the torque about the pivot point (to give a net torque in the direction of rotation and thus passing the step), I'm not able to see what torque can compensate the one from gravity which doesn't let the sphere slide up:

The sphere has an angular velocity, but I'm missing something conceptual because I can't figure out how to formally state that the angular momentum changed the right amount so that the sphere made it past the step.
4. The attempt at a solution
My (possible wrong) attempt was to say that the sphere lost kinetic energy because of the change on potential energy:
\frac{1}{2}I\omega^{2}+\frac{1}{2}mv^2 +mgR = mg(R+h)\\<br /> v=\sqrt{\frac{10}{7}gh}

I hope you can help me :)

 

[Starkrod]

It's v=\sqrt{\frac{10}{7}gh} [\tex]

 

[Hiero]

You are correct in that you are wrong, because as you said, additional energy is lost in the collision.

You say angular momentum changes, but "the angular momentum of an object" does not exist. Only "the angular momentum of an object with respect to an origin" makes sense. Are you sure the angular momentum changes for every choice of origin?

 

[Starkrod]

You are correct in that you are wrong, because as you said, additional energy is lost in the collision.

You say angular momentum changes, but "the angular momentum of an object" does not exist. Only "the angular momentum of an object with respect to an origin" makes sense. Are you sure the angular momentum changes for every choice of origin?

You are suggesting a reference frame in which there's no torque involved? I really don't see it

 

[Hiero]

You are suggesting a reference frame in which there's no torque involved?

Well, not exactly. We typically assume the collision lasts a very short time, such that the gravitational impulse, and the normal impulse from the ground, are ignorable. The torque due to gravity and the lower ground are justifiably ignored in the limit of a small collision time, because the corner force becomes more and more like a dirac delta spike and is thus the dominant effect in that limit.

With that said, yes.

 

[Starkrod]

Well, not exactly. We typically assume the collision lasts a very short time, such that the gravitational impulse, and the normal impulse from the ground, are ignorable. The torque due to gravity and the lower ground are justifiably ignored in the limit of a small collision time, because the corner force becomes more and more like a dirac delta spike and is thus the dominant effect in that limit.

With that said, yes.

Ok, I think I get your argument. If a choose my origin on the pivot I can omit the set up from before and think in two different moments: before the collision I have the sphere with some velocity v and argular momentum \vec{r} x \vec{p}, after the collision I have other velocity and another r and the angular momentum about the pivot is conserved. So I can use conservation of both linear and angular momentum to solve the problem?

 

[Hiero]

What is r? Maybe the position vector of the center of mass with respect to the pivot point? You're saying this changes?

 

[Starkrod]

What is r? Maybe the position vector of the center of mass with respect to the pivot point? You're saying this changes?

I was trying to say r_x thinking two dimensional. Otherwise I feel like out of information.

 

[Starkrod]

I was trying to say r_x thinking two dimensional. Otherwise I feel like out of information.

Or should I think more about the components of that contact force. I'm kind of lost getting information or "critical" relations that satisfy that condition

 

[Hiero]

Well one key relation is that the collision is perfectly inelastic, which we can interpret as meaning that after the collision, the contact point is stuck, meaning it purely rotates about that point after the collision. So we have a ball rolling towards a corner, we know the exact motion of this rolling, and then we know that it purely rotates about that corner after they collide, we just don't know with what angular speed. Well you've pointed out that angular momentum is conserved about that pivot, so, if you are careful, you should be able to determine the angular speed which gives the same angular momentum immediately after the collision as there was before the collision.

Then your energy considerations should help in determining if that angular speed is sufficient to reach the top.

 

[haruspex]

What is the minimum velocity for which the sphere will be over the step?

That is actually a very hard question. Maybe it will rise high enough to be over the step, but peak too soon, and hit the edge of the step again. If it does, and there is friction on impact, will the rotational inertia be enough to complete the ascent in a second trajectory? Or a third??...

In fact, it is not clear whether there is (sufficient) friction on impact with the step the first time to ensure no slipping then. It only says it had been rolling without slipping, which is a tautology.

 

[zwierz]

Assume the ball never slips. What about a reaction force from the floor (at the lowest point of the ball) at the moment of collision?
If the ball just rests on the floor (without motion) and touches the edge of the stair then it is already a statically indeterminate system.

 

[haruspex]

What about a reaction force from the floor (at the lowest point of the ball) at the moment of collision?

I do not see how that could be relevant. As Hiero notes in post #6, all strictly limited forces can be ignored during (what is presumed to be) the very brief impact.

 

[zwierz]

As Hiero notes in post #6, all strictly limited forces can be ignored

why do you think that this reaction must be limited?

 

[haruspex]

why do you think that this reaction must be limited?

Because there is no downward unlimited force threatening to penetrate the floor.

 

[zwierz]

-

 

[zwierz]

downward unlimited force threatening to penetrate the floor.

horizontal forces also play a role in angular momentum equation
and you can never deduce from the equations of impact that these forces are limited ; above I have explained why. If you propose that as a hypothesis -- no problemUPD

 

[Starkrod]

Well one key relation is that the collision is perfectly inelastic, which we can interpret as meaning that after the collision, the contact point is stuck, meaning it purely rotates about that point after the collision. So we have a ball rolling towards a corner, we know the exact motion of this rolling, and then we know that it purely rotates about that corner after they collide, we just don't know with what angular speed. Well you've pointed out that angular momentum is conserved about that pivot, so, if you are careful, you should be able to determine the angular speed which gives the same angular momentum immediately after the collision as there was before the collision.

Then your energy considerations should help in determining if that angular speed is sufficient to reach the top.

Man, thank you very much. I asked in other sites and that seems to be a satisfying answer, I was trying to think of it as an statics problem and got really confused. I guess the word collision is key to all the considerations you pointed out.

That is actually a very hard question. Maybe it will rise high enough to be over the step, but peak too soon, and hit the edge of the step again. If it does, and there is friction on impact, will the rotational inertia be enough to complete the ascent in a second trajectory? Or a third??...

In fact, it is not clear whether there is (sufficient) friction on impact with the step the first time to ensure no slipping then. It only says it had been rolling without slipping, which is a tautology.

It was a hard question for me too, because yes, it is open to suggest the right conditions for the sphere to go up, not only it's velocity and it happens in a lot of these classical mechanics problems from old Russian books.