B Will a round-headed rod topple if it slides down a frictionless slope?

[Baluncore]

Cancel all friction, then place one coin on the slope, and watch it accelerate away down-slope.

Place one coin on top of another, then place the two on the slope. The two coins, with one resting on the other, will accelerate together down-slope, with identical velocities.

A stack of coins will not tumble, it will move as one cylinder down-slope. The stack of coins is a virtual rod, that is standing perpendicular to the slope.

The rod need not have a rounded foot. Like a coin, it could have a flat foot.

 

[Ibix]

Why do we consider torque only around center of mass?

You can consider moments around the end of the rod if you like, but then you need to include the $ma$ of the rod's center of mass accelerating downslope, which I didn't draw (it has no moment about the center of mass so it doesn't matter to my earlier argument). Since $ma=mg\sin\theta$, the "rotation" of the rod around its end is just the rate of change of its angular coordinate as it slides down the slope.

 

[jbriggs444]

I agree that a rod, perpendicular to slope, with a rounded tip will slide without tipping. Depending on the curvature of the rod's rounded tip, this orientation may be stable or unstable.

A vertical rod with a rounded tip can slide without tipping if the center of curvature coincides with the rod's center of gravity and the slope is gentle enough that the contact point stays somewhere in the interior of the rounded tip.

A vertical rod with a rounded tip can slide and tip toward a stable perpendicular orientation if the center of curvature is beyond the rod's center of gravity and if the slope is sufficiently gentle to keep the contact point in bounds. The rod will rock back and forth about the perpendicular in this case.

Like one of these.

Edit to add one more case...

If center of curvature of the tip of a non-perpendicular rod is on the near side of the center of gravity then instead of rocking toward the perpendicular, the rod will topple away from the perpendicular. This is likely the case that most of us have in mind when we think about a rod with a rounded tip on an slippery inclined slope.

 

[Baluncore]

I agree that a rod, perpendicular to slope, with a rounded tip will slide without tipping. Depending on the curvature of the rod's rounded tip, this orientation may be stable or unstable.

You seem to be assuming there is some friction present at the contact patch.
Without friction, if the contact patch, contains the perpendicular through the centre of gravity, then the shape of the rod-end, and of the contact patch is quite irrelevant.

 

[Delta2]

Guys (@jbriggs444 and @Baluncore) when you say vertical rod you mean with its axis parallel to gravity, while when you say perpendicular to the slope you mean with its axis perpendicular to the slope right?

 

[jbriggs444]

You seem to be assuming there is some friction present at the contact patch.

Not at all. Quite the opposite. Without friction, the perpendicular orientation is an equilibrium point regardless of curvature. With friction, it is (almost certainly) not. We agree on that.

If one is to consider stability, the curvature of the surface near the contact point becomes relevant, even without friction. Both a needle and a bopping buddy are in equilibrium in a perpendicular orientation. The needle is unstable at that equilibrium. The bopping buddy is stable there.

A single coin in a "perpendicular" orientation (flat on its face) will be in a stable equilibrium with or without friction [barring a rather extreme slope and coefficient of friction].

 

[jbriggs444]

Guys (@jbriggs444 and @Baluncore) when you say vertical rod you mean with its axis parallel to gravity, while when you say perpendicular to the slope you mean with its axis perpendicular to the slope right?

Yes. By vertical I mean aligned with gravity and by perpendicular I mean perpendicular to the slope.

 

[phinds]

This is a graphic of my original analysis posted in #8 and #10. We've gone 'round this so much that I now can't tell who agrees w/ me and who doesn't and why.

SO ... if this is wrong, can someone explain to me WHY it is wrong and if it's right can we all just agree to stop going 'round and 'round?

 

[Baluncore]

SO ... if this is wrong, can someone explain to me WHY it is wrong and if it's right can we all just agree to stop going 'round and 'round?

I agree, but you are ignoring the special case of a rod perpendicular to the slope.

Now, replace the ball with a round headed rod, does this means that the rod won't topple no matter the rod is perpendicular to the slope or not?

The perpendicular rod will not topple.

 

[phinds]

I agree, but you are ignoring the special case of a rod perpendicular to the slope.

I don't care about that case, I'm only dealing w/ the vertical rod. In case anyone still has doubts, let me give the full vector analysis of the vertical rod, extending what I said in post #38

This is just the vertical special case of the more general solution that @Ibix presented in post #26

 

[Delta2]

don't care about that case,

In my opinion you should care, at least me finds it very counterintuitive that the perpendicular to the slope rod will not topple

 

[Baluncore]

I don't care about that case, I'm only dealing w/ the vertical rod.

I do care. The vertical rod is a trivial and obvious case, it will topple.

The perpendicular rod is the core of the sphere, and follows the same physics as the sphere. A sphere will not roll, but will slide, a perpendicular rod will not topple.

 

[alan123hk]

The shape of the bottom of the object in contact with the sloped surface has a big impact. For example, as shown in the image below, A will obviously topple, while B will most likely not.

​

 

[DaveC426913]

Place one coin on top of another, then place the two on the slope. The two coins, with one resting on the other, will accelerate together down-slope, with identical velocities.

(I am still struggling with frictionless surfaces.)

The coin in contact with the slope will be pushed laterally by its encounter with the slope - frictionless or otherwise - will it not? It does acquire some amount of lateral acceleration.

So what equivalent lateral force would act on the top penny to start it moving laterally (as opposed to just getting left behind by the bottom penny)?

 

[Baluncore]

So what equivalent lateral force would act on the top penny to start it moving laterally (as opposed to just getting left behind by the bottom penny)?

The top coin slides off the bottom coin, at the same speed as the bottom coin slides down the slope.

The top coin slides at the same rate as the bottom coin, because they are both subjected to the same acceleration, and there is no friction between them.

 

[DaveC426913]

The top coin slides off the bottom coin, at the same speed as the bottom coin slides down the slope.

The top coin slides at the same rate as the bottom coin, because they are both subjected to the same acceleration, and there is no friction between them.

I still don't get it. The bottom coin is subjected to a lateral force; the top coin is not.

[EDIT] Oh now I get it. We're talking about the rod that's perpendicular to the slope. The top penny is pushed along by the bottom penny, which is behind it.

 

[James Brown]

The perpendicular rod will not topple.

Which means that if the rod is placed like that:(figure below), the rod won't topple?

If the force that act on the rod is mg, the rod will topple right?, but if the force that act on the rod is mgsinx then it will make sense that the rod won't topple

 

[Ibix]

If the force that act on the rod is mg, the rod will topple right?, but if the force that act on the rod is mgsinx then it will make sense that the rod won't topple

Forces are vectors. The weight force has magnitude $mg$, but the component parallel to the slope is $mg\sin x$, assuming $x$ is the angle the slope makes with the horizontal.

 

[Baluncore]

Which means that if the rod is placed like that:(figure below), the rod won't topple?

You can balance a pin on its point for a while, but it will eventually fall. The same will happen to the perpendicular rod, but not before it slides off the end of the ramp. All the parts of the rod are accelerating at the same rate and there is no friction to start a rotation.

 

[jbriggs444]

This is a graphic of my original analysis posted in #8 and #10. We've gone 'round this so much that I now can't tell who agrees w/ me and who doesn't and why.

View attachment 302680
SO ... if this is wrong, can someone explain to me WHY it is wrong and if it's right can we all just agree to stop going 'round and 'round?

If the slope is shallower than depicted and if the rod is rounded off, with a center of curvature up above the COG then the line of action of that normal force can be on the other side of the COG.

 

[Baluncore]

I still don't get it. The bottom coin is subjected to a lateral force; the top coin is not.

[EDIT] Oh now I get it. We're talking about the rod that's perpendicular to the slope. The top penny is pushed along by the bottom penny, which is behind it.

The coins are subjected to identical forces, because there is no friction.
The top penny is resting flat on the bottom penny, which is below it in the stack of coins.

 

[Baluncore]

The shape of the bottom of the object in contact with the sloped surface has a big impact. For example, as shown in the image below, A will obviously topple, while B will most likely not.

Again; You can balance a pin on its point for a while, but it will eventually fall. Likewise, you can stand a rod on a horizontal surface, and it can be stable if it has a sufficiently flat end-profile.

But without friction, a rod would slide off the end of the ramp before it began to lean away from the perpendicular. In the short term, the shape of the rod-end in contact with the ramp is irrelevant, so long as a perpendicular to the ramp, that passes through the centre of gravity, also passes through the contact patch or point.

 

[DaveC426913]

The coins are subjected to identical forces, because there is no friction.

With all due respect, that's not true. There is friction between the coins.

The top penny is resting flat on the bottom penny, which is below it in the stack of coins.

We need clarification of these terms.

Define 'flat'. Define 'below'. WRT gravity or WRT to the slope?

See, if WRT gravity (which is what I think you mean), then the bottom coin experiences a lateral force (even from a frictionless surface). We know this is true because it begins to accelerate laterally as it falls (if it did not accelerate laterally, it would have to penetrate the slope vertically).

Again, if the top penny is resting on the bottom penny "flat" wrt to gravity then it does not experience any lateral force. Thus it will fall straight down.

 

[jbriggs444]

Absent friction, one can adopt a coordinate system that is rotated to match the slope of the plane and which accelerates down-slope at the same rate that an object would slide.

Adopting this as a frame of reference, one has an ordinary, slippery, flat, horizontal surface with a reduced vertical acceleration of gravity. A rod which is perpendicular to the inclined surface is vertical in this frame. Its motion can be analyzed accordingly.

 

[Baluncore]

With all due respect, that's not true. There is friction between the coins.

The slope is accepted to be friction free. I hereby define the coins as also being friction free.

Define 'flat'. Define 'below'. WRT gravity or WRT to the slope?

The slope is not horizontal, but it is flat. The coin is flat on the surface of the slope. Flat with reference to the slope.

Imagine a stack of coins on a level horizontal surface, then tilt that surface with the coins to the slope in question. Next, switch off all friction between the slope and between individual coins, to watch the stack of slippery coins slide off the end of the ramp, without toppling during the process.

 

[James Brown]

So the force that cause the rod to topple when there is no friction is the reaction force acting on the rod right?

 

[Delta2]

So the force that cause the rod to topple when there is no friction is the reaction force acting on the rod right?

If by reaction force you mean the normal force from the slope then yes I think so.

 

[Steve4Physics]

May or may not help, but a few thoughts in addition to what has already been said...

First consider why a ball (released from rest say) on a frictionless slope slides without rolling.

The two forces on the ball (its weight, W and the normal reaction, R from the slope) both act through the ball’s centre of gravity (C). So there is no net torque about C.

(There is a torque about the point of contact (P) of the ball and plane. because W does not pass through P. But this does not lead to rolling- e.g. see https://www.physicsforums.com/threads/can-a-ball-roll-down-a-frictionless-plane.271692/)

For the rounded rod (or any other shape with centre of gravity C and a single point of contact, P, with the plane), if CP is normal (perpendicular) to the slope, there is no torque about C because both R and W pass through C. So there is no angular acceleration about C - the rod will not rotate as it slides down.

But if CP is not normal (not perpendicular) to the slope, then R produces a torque about C. (It’s probably better to think in terms of a couple: there is a couple - formed by R and a component of W, acting on the rod.) So the rod rotates (‘falls over’) while it slides downhill.

 

[SammyS]

I don't care about that case, I'm only dealing w/ the vertical rod. In case anyone still has doubts, let me give the full vector analysis of the vertical rod, extending what I said in post #38

View attachment 302681

This is just the vertical special case of the more general solution that @Ibix presented in post #26

This seems like a reasonable point to make. After all, the statement of the problem includes the following (Underlining added for emphasis.):

Now, replace the ball with a round headed rod, does this means that the rod won't topple no matter the rod is perpendicular to the slope or not?

 

[jrmichler]

The diagram in Post #43 is not correct. Here is the free body diagram of a pointed rod perpendicular to a sloped frictionless surface at angle $\alpha$. The gravity force $mg$ is vertically downward. It is decomposed into components parallel to the rod and perpendicular to the rod. The gravity force component against the slope is $mg \cos \alpha$, and is perpendicular to the frictionless surface. The gravity force component parallel to the slope is $mg \sin \alpha$. That force component accelerates the rod down the slope. It is counterbalanced by the acceleration force $ma$.

All forces sum to zero, moments sum to zero, the rod slides down the slope without tipping over. This is identical to the case of a ball sliding down a frictionless slope. It is easier to visualize if you think of a ball sliding down the slope, then remove the parts of the ball that are not a pointed rod.