1
$\begingroup$

This is a 1998 IPHO Mechanics problem analyzing a rolling hexagonal prism (a pencil). A similar question was asked on PSE but the question is unanswered and nobody gave a proper explanation but just a link to the answer solution.

enter image description here

enter image description here

The solution said that the angular momentum about the vertex that hit the plane after the collision is conserved during the brief period of impact because the torque done by the force is zero because the perpendicular distance is zero. The reason makes sense, but I don’t see how net torque is zero as this is the requirement for the conservation of angular momentum. Isn’t there torque done by gravity?

$\endgroup$
2

2 Answers 2

1
$\begingroup$

I have added a few labels to a diagram given in the solution to this problem.

enter image description here

The prism has three forces acting on it, a normal force and a frictional force at the point of contact, and a gravitational force, $mg$.

Finding the angular momentum about a point of contact means that only the torque due to the gravitational force needs to be considered being the only external torque about a point of contact, $\tau$, acting on the prism.

The motion described has the prism rotating about $A$ followed by a rotation about $B$.

During the time of transition from rotation about $A$ to rotation about $B$, $\Delta t$, an impulsive torque $\int_0^{\Delta t} \tau\,\mathrm{d}t$ acts on the prism.

This impulsive torque can be neglected on the assumption that $\Delta t \to 0$, i.e., the transition, $v_{\rm A}$ to $v_{\rm B}$, occurs (almost) instantaneously, and if that is the case then angular momentum is conserved about any position on the slope.

All well and good?

However, you might consider doing the analysis by looking at the change in angular momentum about the centre of mass $C$. That change in angular momentum is effected by an external torque due to the contact forces on the prism due to the slope. Why does the impulsive torque due to those contact forces have an effect on the angular momentum whereas the impulsive torque due to gravitational attraction in the previous analysis could be neglected?

And the answer is a matter of scale in that the contact forces during this phase of the motion are much greater than the gravitational force. In the real world the gravitational force does contribute to an impulsive torque as there is a finite transition time but that impulsive torque is very much smaller than the contact forces impulsive torques.

Related to this is the answer to the question, Can linear momentum be conserved before and after collision in the presence of an external force?.

$\endgroup$
1
$\begingroup$

The solution said that the angular momentum about the vertex that hit the plane after the collision is conserved during the brief period of impact because the torque done by the force is zero because the perpendicular distance is zero.

If this refers to the gravitational or downhill force, this argument is wrong. First, we can summarise the gravitational force and the resulting torque as acting on the centre of mass, which rests in the centre of the hexagon. For the perpendicular distance to be zero, the force would have to point towards the edge. Instead it points in the direction of plane (due to the constraints imposed by the ground).

It may help to look at the limit of an infinite number of edges, i.e., a cylinder rolling down the plane. If gravity wouldn’t enact any torque on it, it would never accelerate.

I don’t see how net torque is zero as this is the requirement for the conservation of angular momentum. Isn’t there torque done by gravity?

The reason why the torque caused by gravity can be ignored is that we are looking at an instantaneous collision and the torque only contributes over an extended time. If you so wish, you can switch off gravity for a moment around the impact and nothing would change.

Coming at it from the other side: For the conservation of angular momentum to be broken, some angular momentum needs to be added to the situation. Torque acting over time (or angle) results in angular momentum, but since we are looking at an instantaneous event, our gravitational torque doesn’t get to do that.

You might compare this to a ball bouncing off a wall. If you just look at its state directly before and after the impact, the gravitational force doesn’t matter – you don’t even need to know in which direction it’s pointing. It requires the ball to traverse some distance to affect its momentum. Your situation is quite similar, just with all the linear quantities being replaced by rotational ones.

$\endgroup$
4
  • $\begingroup$ sry, my bad. When I said perpendicular distance is zero I am talking about the torque produced by the Force when the vertex hits the incline, that instantaneous force, i don't mean perpendicular distance between force of gravity and the vertex. $\endgroup$
    – Hammock
    Oct 31, 2023 at 1:03
  • $\begingroup$ @Hammock: Well, this might have been avoided if you had linked to the solution in question … $\endgroup$
    – Wrzlprmft
    Oct 31, 2023 at 7:33
  • $\begingroup$ @Wrzlprmft Careful about switching off gravity as the normal and frictional forces will also be switched off? $\endgroup$
    – Farcher
    Oct 31, 2023 at 10:34
  • $\begingroup$ @Farcher: Those forces also only have a possible impact over time. For the collision, the only thing that matters is the interaction between the surfaces of two rigid objects, namely the roller and the ground. (So, it’s electromagnetic if you go down to the bottom of it.) $\endgroup$
    – Wrzlprmft
    Oct 31, 2023 at 11:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.