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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?

enter image description here

Relevant 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

$v=\omega R$

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 which torque can overcome the one from gravity which doesn't let the sphere slide up:

enter image description here

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.

The attempt at a solution

My (almost certainly 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)\\ v=\sqrt{\frac{10}{7}gh} $

I hope you can help me :)

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closed as off-topic by Kyle Kanos, Yashas, ZeroTheHero, Jon Custer, David Z Jun 12 '17 at 2:04

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  • $\begingroup$ Are you sure the collision is inelastic? $\endgroup$ – probably_someone Jun 11 '17 at 4:17
  • $\begingroup$ The hard part is figuring out how the sphere is moving immediately after it contacts the step. It isn't stated clearly. The velocity of the point in contact is $0$. Suppose the step was taller than the sphere. Is the problem saying the sphere would lose all forward velocity, but keep rotating at the same speed? Since it does not slip, it would then acquire a vertical velocity and roll up the wall. $\endgroup$ – mmesser314 Jun 11 '17 at 4:24
  • $\begingroup$ Your solution assumes that no KE is lost during the collision with the step. However, as is stated in the question, this collision is completely inelastic : the relative velocity between the sphere and the step normal the the common surface becomes zero. So some KE is lost before the sphere begins to roll about the point of contact. $\endgroup$ – sammy gerbil Jun 11 '17 at 8:14
  • $\begingroup$ Yes I realized that, I just wasn't thinking the problem the right way, the word collision has a lot considerations that my "static" approach doesn't considerate. $\endgroup$ – David Leonardo Ramos Jun 11 '17 at 12:57
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    $\begingroup$ The question may be a homework, but it is high quality and interesting. $\endgroup$ – user259412 Jun 11 '17 at 15:40
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Angular momentum can be conserved but only under a certain condition.

The angular momentum just before collision about a point = angular momentum just after collision about the same point.

Hence,we shall conserve angular momentum about the tip of the elevation(as shown in second picture).

Anugular momentum about external point in combined translation and rotation = $L_{com} + mvr Sin\theta$ $$L_1 = mv(R-h)+I\omega$$ $$L_2 = (I+mR^2)\omega_{2}$$ $$L_1 = L_2$$ also , $\omega = vR$ (pure rolling)

$$mv(R-h)+I\omega = (I+mR^2)\omega_2$$ $$\omega_2 = \frac{mv(R-h)+I\omega}{I+mR^2}$$

Now, Kinetic energy just after collision should be just enough to elevate the sphere by h.

$$\frac{1}{2}(I+mR^2)\omega_2^2 = mgh$$

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  • $\begingroup$ Thank you I was having a hard time with the conservation of angular momentum because in a static set up there is torque. But you and @Farcher made it clear that I do have to think of it as a collision and can ignore that torque. Also in the last equation should be $... = mg(R+h)$ right? $\endgroup$ – David Leonardo Ramos Jun 11 '17 at 13:52
  • $\begingroup$ My pleasure, David. Although I believe that the last equation should still be mgh, as the initial potential energy was mgR and the final potential energy is mg(R+h), and thus the difference will be mgh. $\endgroup$ – HyperBean Jun 11 '17 at 15:38
  • $\begingroup$ Yeah I missed the missing potential energy on the left. $\endgroup$ – David Leonardo Ramos Jun 11 '17 at 15:45
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You do need to use the conservation of angular momentum and the conservation of energy but you need justification for their use.

When the sphere hits the step there are three forces acting on it.

  • The weight of the sphere
  • The normal reaction of the step on the sphere.
  • The force of friction on the sphere.

In terms of looking at the torques about the point of contact of the sphere and the step acting on the sphere you had no difficulty in eliminating the normal force and the friction force in that they do not exert a torque about the point of contact.

As with a lot of collision type problems what you have to realise is that the normal forces and the friction force are impulsive forces whereas the weight is not.
So the assumption which is made is that the "collision" occurs over a very short period of time and the impulsive forces are very much larger than the weight of the sphere.
This in turn means that the impulse due to the weight over the period of the "collision" is very small and can be neglected.
I have tried to show this here.

So because the "collision" between the step and the sphere takes place over such a short period of time the impulsive torque due to the weight of the sphere can be therefore neglected and conservation of angular momentum can be used about the point of contact.

Immediately after the "collision" between the sphere and the step you assume that there is no slipping between the sphere and the step so the normal reaction force and the frictional force between the sphere and the step do no work as those forces do not move.
So the only force which you need to consider is the weight of the sphere acting at the centre of mass of the sphere.
You can then use the conservation of energy to equate the work done by the weight (or the change in gravitational potential energy of the sphere and earth system) to the change in kinetic energy of the sphere.

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  • $\begingroup$ Doesnt the friction force between the sphere and step aplly impulsive torque and prevent us from using comservation of angular momentum? $\endgroup$ – alex Jun 11 '17 at 8:20
  • $\begingroup$ The frictional force acts at the point about which the torque is calculated, the top of the step. $\endgroup$ – Farcher Jun 11 '17 at 8:21
  • $\begingroup$ oh sorry my bad. $\endgroup$ – alex Jun 11 '17 at 10:08
  • $\begingroup$ Thank you very much, your answer compiled all that I was confused about when thinking about the problem. There was a lot of conditions to have in mind because, as you pointed out, you have to assume also that there is no slipping after the collision which for me is a wild guess. Thank you again. $\endgroup$ – David Leonardo Ramos Jun 11 '17 at 14:02

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