How does angular momentum conservation apply to a cylinder rolling up a step?

[haruspex]

My thinking is that the angular momentum about the step just before collision is zero, because the center of mass of the cylinder is moving directly towards the edge of the step.

The cylinder's rotation on its own axis also constitutes angular momentum about the step corner. To make this obvious, consider the top and bottom halves of the cylinder separately - the top half is moving faster towards the step so has greater moment about the step then the lower half has in the other rotational direction.

 

[rcgldr]

The cylinder's rotation on its own axis also constitutes angular momentum about the step corner.

OK, now I understand that part, but to make this a closed system, the angular and linear momentum of whatever the step is attached to, for example the earth, also needs to be included, if linear and angular momentum are to be conserved. I don't know if energy is conserved because the problem doesn't state that the collision is elastic.

 

[haruspex]

OK, now I understand that part, but to make this a closed system, the angular and linear momentum of whatever the step is attached to, for example the earth, also needs to be included, if linear and angular momentum are to be conserved. I don't know if energy is conserved because the problem doesn't state that the collision is elastic.

Angular momentum is only meaningful in the context of a reference point. We're only considering the angular momentum of the cylinder itself, taken about the step corner as reference. Since the impulse from the step corner has no moment in that context, the angular momentum of the cylinder about it is conserved.
Wrt energy conservation, the relevant statement in the OP is "Assuming cylinder remains in contact". This has to be interpreted as 'transiently' since in fact it will become airborne. But it does mean there is no bounce, so the collision should be taken as inelastic.

 

[ehild]

Wrt energy conservation, the relevant statement in the OP is "Assuming cylinder remains in contact". This has to be interpreted as 'transiently' since in fact it will become airborne. But it does mean there is no bounce, so the collision should be taken as inelastic.

See URL.

http://en.wikipedia.org/wiki/Work_(physics)

In physics, a force is said to do work when it acts on a body, and there is a displacement of the point of application in the direction of the force.

The force from the wedge has two components: one radial, that causes deformation, elastic or inelastic. The other component is tangential: the (static) force of friction. It does no work.

"No bounce" is not among the requirements of an elastic collision. If a perfectly elastic ball hits an other identical one initially at rest, the hitting ball becomes stationary and the other one will move further. That is an elastic collision, and there is no bounce.
Rutherford scattering is also considered "elastic collision" and the interacting atoms change direction, but there is no bounce.

In he problem, the linear momentum of the CM changes direction and its rotation changes angular velocity but it moves with respect to the wedge during the whole interaction. Kinetic energy can be conserved, so the collision can be elastic .


ehild

 

[haruspex]

The force from the wedge has two components: one radial, that causes deformation, elastic or inelastic. The other component is tangential: the (static) force of friction. It does no work.

In an elastic collision with a tangential (frictional) component, elastic deformation also occurs in the tangential direction. When a superball spinning on a horizontal axis is dropped onto a hard surface its spin reverses.

"No bounce" is not among the requirements of an elastic collision. If a perfectly elastic ball hits an other identical one initially at rest, the hitting ball becomes stationary and the other one will move further. That is an elastic collision, and there is no bounce.

But in this case the other object, the step, is fixed.

Rutherford scattering is also considered "elastic collision" and the interacting atoms change direction, but there is no bounce.

How do you know?

In the problem, the linear momentum of the CM changes direction and its rotation changes angular velocity but it moves with respect to the wedge during the whole interaction. Kinetic energy can be conserved, so the collision can be elastic .

Consider the direction of movement of the cylinder's CoM immediately after collision. At one extreme, it will move in a line perpendicular to the radius to the point of contact with the step; it cannot take any lower line or it would move into the step. If there were any bounce at all it would take a higher line (consider velocity along that radius and coefficient of restitution). The statement given is that it 'remains in contact with the step without slipping'. Though the first part of that is not strictly correct, I take it to mean that immediately after the collision the point on the cylinder that collided with the step has zero velocity. That establishes the CoR is zero, and the friction is enough to prevent slipping.
Fwiw, I calculate the minimum coefficient of static friction to prevent slipping to be 3 sin(θ)/(1-cos(θ)), where the step height is r(1-cos(θ)).

 

[ehild]

Collision is elastic if the kinetic energy is the same before and after the collision. During the collision, part of the mechanical energy is elastic.
When a ball collides to a stationary wall the normal component of the velocity of its centre becomes zero for an instant - but the collision might be elastic, if the deformations during the collision are elastic.

The cylinder experiences both a normal force and a tangential one during the interaction, when it is in contact with the step, and their resultant force determines how the centre moves. There are deformations during the collision, and the step and cylinder have a contact area, not only a contact point. The collision lasts till the cylinder is in contact with the step. The centre can have a radial velocity at the end and loosing contact - but it is not the collision process any more. The problem does not say that the cylinder stays in contact with the step after the collision.

If we accept that the cylinder remains in contact with the step (no radial component of the velocity of CM) and rotates about it as fixed point even after the collision, the final angular velocity is 5/6ω₀, and the final and initial kinetic energies really are not the same.

ehild

 

[TheTank]

Did no one come up with a solution to the problem? I am trying to solve this problem as well.. My book says the answere should be, if the velocity of the cylinder (c.o.m) is v, then it will have a new velocity v' = v/3 (1+2cos(angle of impact)) shortly after impact. Hope this helps someone to make an attempt..

 

[haruspex]

Did no one come up with a solution to the problem? I am trying to solve this problem as well.. My book says the answere should be, if the velocity of the cylinder (c.o.m) is v, then it will have a new velocity v' = v/3 (1+2cos(angle of impact)) shortly after impact. Hope this helps someone to make an attempt..

See post #5.

 

[ritajit]

can someone please explain the meaning of the term 3mvr/4 in the solution?

 

[haruspex]

can someone please explain the meaning of the term 3mvr/4 in the solution?

The initial angular momentum about the step edge can be thought of as made of two parts:
- the angular momentum due to the cylinder's rotation about its centre, plus
- the angular momentum contribution from its linear motion.
For the second, its linear momentum is mv. The height of its mass centre above the step edge is 3r/4. That height is orthogonal to the linear motion, so the angular momentum from its linear motion is the product of the two: 3mvr/4 = 3mr²ω/4.
Adding the rotation about its centre we get 3mr²ω/4+mr²ω/2=5mr²ω/4.

 

[ritajit]

The initial angular momentum about the step edge can be thought of as made of two parts:
- the angular momentum due to the cylinder's rotation about its centre, plus
- the angular momentum contribution from its linear motion.
For the second, its linear momentum is mv. The height of its mass centre above the step edge is 3r/4. That height is orthogonal to the linear motion, so the angular momentum from its linear motion is the product of the two: 3mvr/4 = 3mr²ω/4.
Adding the rotation about its centre we get 3mr²ω/4+mr²ω/2=5mr²ω/4.

Thanks a lot, that was fast.
One more thing, I always thought that while using the law of conservation of angular momentum you should only keep it about a single point. Rotational Dynamics is my weakness so I'd be really grateful if you could shed some light on this.
Thanks again :)

 

[haruspex]

Thanks a lot, that was fast.
One more thing, I always thought that while using the law of conservation of angular momentum you should only keep it about a single point. Rotational Dynamics is my weakness so I'd be really grateful if you could shed some light on this.
Thanks again :)

You can take it about any fixed point as long as you are consistent. You can also take it about the mass centre of the rigid body, as a moving reference axis. The instantaneous centre of rotation counts as a fixed point for this purpose. Taking it about any other point which is fixed relative to the rigid body is generally invalid.
See section 5 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-moments/

 

[True]

How can we ignore the torque due to friction in this question?

 

[haruspex]

How can we ignore the torque due to friction in this question?

If we take moments about the point of contact, neither the normal force/impulse nor the frictional force/impulse has any moment about it.