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

[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.

 

[Melbourne Guy]

Does it help to approach this from the perspective of dropping a rod on the Moon? It falls in whatever orientation it started in, with no change in orientation, to the surface. That's essentially a frictionless situation. Add in a slope of any kind, and if it is also frictionless, surely the outcome is the same?

 

[alan123hk]

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.

I think I understand your argument and I think your argument makes sense. But on the other hand, my intuition tells me that even if there is no friction on the slope surface, the situation might look like the picture below.

​

Honestly, I'm still not sure about the answer to this question. This seemingly simple problem seems to be more complicated than imagined.

 

[jbriggs444]

surely the outcome is the same?

If you drop a can of Tinker Toys or pick-up sticks on the floor, they do not normally all wind up standing on end. While I have not tried that experiment on an inclined plane in many years, I have little doubt about the outcome.

 

[Delta2]

@alan123hk I think the torque of weight around a point in the contact base is something like a red herring. All that it matters is the torque of the normal force.

 

[Melbourne Guy]

If you drop a can of Tinker Toys or pick-up sticks on the floor, they do not normally all wind up standing on end.

Is that what's been described in the OP, though, @jbriggs444? Isn't the question about the motion as it falls, not the end state?

 

[hutchphd]

The friction is either zero or not. Zero means 0.

 

[jbriggs444]

Is that what's been described in the OP

You were the one who first talked about free fall in vacuum without a surface.

Isn't the question about the motion as it falls, not the end state?

The motion as it falls is a transition from initial state to end state. If it winds up flat on the floor, it probably toppled to get there.

 

[Bystander]

If it winds up flat on the floor, it probably toppled to get there.

"You can't fall off the floor." ---W.C. Fields; probably not, but the attribution is apropos.

 

[Melbourne Guy]

You were the one who first talked about free fall in vacuum without a surface.

Yes, to try and visualise the situation. The OP did not say air or airless. If there is air, then there is a frictional force applied, but this seems an idealised question because there are no frictionless surfaces, so idealised, there is no air. I do not see, if the rod does not topple in free fall, how a frictionless surface adds a force that causes torque. But it you add air resistance, then fine, it can topple.

 

[James Brown]

Ok so, does this means that the force that the rod to rotate is mg while the force that cause the rod to accelerate downward is mgsinx right?

 

[Delta2]

I do not see, if the rod does not topple in free fall, how a frictionless surface adds a force that causes torque

I am not an expert in rigid body dynamics but it seems to me that the torque of the normal force from the frictionless surface is one of the key things here.

 

[DaveC426913]

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

This is where we disagree. Your inference is unwarranted in my view. At the very least, I think it behooved you to state it explicitly in your assumptions.

But I think, in general, everyone has been doing a pretty poor job of defining the problem - and solutions - clearly. This thread would have been about 20 posts shorter.

 

[Baluncore]

This is where we disagree. Your inference is unwarranted in my view.

That is not an inference, it was the original definition for my model of a rod as being a stack of coins. A slope or a real coin does have friction, imaginary slopes and coins can be defined to be friction free. Have you never wondered why money slides so easily through your fingers?

 

[DaveC426913]

...imaginary slopes and coins can be defined to be friction free.

But they were not. The slope was.

If your solutions depended on defining the coins to be friction-free, you needed to state it. It could have saved a lot of confusion.

 

[Baluncore]

If your solutions depended on defining the coins to be friction-free, you needed to state it. It could have saved a lot of confusion.

Post #31

Cancel all friction, ...

 

[alan123hk]

Hopefully someone can help with an experiment, which is to put objects of different shapes at different angles on a slope with very little friction, shoot and upload a video with high resolution and frame rate for our reference.

 

[Melbourne Guy]

...but it seems to me that the torque of the normal force from the frictionless surface is one of the key things here.

What is the 'normal force', @Delta2? Are you referring to a force imparted by the slope?

 

[Delta2]

What is the 'normal force', @Delta2? Are you referring to a force imparted by the slope?

The slope can exert two kind of force: Parallel to its surface, which is friction, and perpendicular to its surface which is called the normal force. The normal force cancels the mgcosθ component of the weight and makes the body's trajectory to follow the incline instead of doing free fall.

 

[jbriggs444]

What is the 'normal force', @Delta2? Are you referring to a force imparted by the slope?

The "normal force" is the contact force between an object and a surface that acts perpendicular to the surface. The word "normal" here actually means "perpendicular".

The normal force exists because objects cannot normally interpenetrate. The normal force is the manifestation of this fact. It is the force that keeps your feet from falling through the floor. It allows balls to bounce off the walls. It allows your fingers to push the buttons on your keyboard.

Yes, it is a force imparted by the slope.

 

[Melbourne Guy]

Yes, it is a force imparted by the slope.

Hypothetically, can a frictionless surface impart the normal force? That's the part I admit to struggling with.

 

[Delta2]

Hypothetically, can a frictionless surface impart the normal force? That's the part I admit to struggling with.

Yes because the friction and the normal force don't have common component, because they are perpendicular to each other.

I don't think its the right place here to open a can of worms regarding the microscopic nature of the friction or the normal force and whether normal force force is due to EM interaction or due to the pauli exclusion principle.

 

[Melbourne Guy]

Yes because the friction and the normal force don't have common component, because they are perpendicular to each other.

Then, in the OP's scenario, the normal force will provide a mechanism for the rod to tumble? Is my understanding correct?

I don't think its the right place here to open a can of worms regarding the microscopic nature of the friction or the normal force and whether normal force force is due to EM interaction or due to the pauli exclusion principle.

That's fine, @Delta2, neither do I. I'm just trying to understand what forces are in play and which ones can be ignored

 

[Lnewqban]

Hypothetically, can a frictionless surface impart the normal force? That's the part I admit to struggling with.

The tires of a hydroplaning car receive exactly the same vertical reaction force from the horizontal pavement than when the car is parked on it.

 

[Delta2]

Then, in the OP's scenario, the normal force will provide a mechanism for the rod to tumble? Is my understanding correct?

Yes exactly. The only case that the torque of normal force is zero (and hence the rod won't tumble) is that the rod is exactly perpendicular to the incline.

 

[Melbourne Guy]

The tires of a hydroplaning car receive exactly the same vertical reaction force from the horizontal pavement than when the car is parked on it.

Isn't that the heart of the OP's question, @Lnewqban? If there is no force to rotate the tyres, doesn't that mean there is no force to topple the rod? Essentially, only gravity is acting on the rod, irrespective of the shape of the tip?

 

[Lnewqban]

Isn't that the heart of the OP's question, @Lnewqban? If there is no force to rotate the tyres, doesn't that mean there is no force to topple the rod? Essentially, only gravity is acting on the rod, irrespective of the shape of the tip?

I agree with you.
The frictionless slope is not adding any force that could induce a moment, if the center of mass is aligned with the direction of that normal force.

 

[alan123hk]

Other examples are as follows. Even if the surface is completely free of any friction, IMHO I think it's clear that the long rods will still fall down.

 

[Melbourne Guy]

Other examples are as follows. Even if the surface is completely free of any friction, IMHO I think it's clear that the long rods will still fall down.

But isn't every part of the rod being acted on by the same gravitational force, @alan123hk? Especially in your second diagram. If friction does not impart any force on the contact point, why would it move any faster or slower than the other end?

 

[alan123hk]

Especially in your second diagram. If friction does not impart any force on the contact point, why would it move any faster or slower than the other end?

This needs to be proven mathematically. I believe a mathematical analysis of the initial state of the system might be helpful, as a first step we can try to simply compare the initial downward acceleration on the left and right ends. If I have time I will try it.

Edit : Think again, it feels like if the inclination is small, the long rod will tip over like in the second diagram, but if the inclination is large, it is very different, I don't know what will happen yet, maybe it will tip over slowly or never tip over, but of course the long rod in the horizontal state will accelerates to the left and down..

 

[Steve4Physics]

It’s interesting (maybe!) to note that you don’t actually need a frictionless surface for the rod to slide down without rotating.

Suppose there is some friction and the rod is ‘leaning backwards’ as it slides down (angle α between slope and rod). For some value of α, the resultant of the frictional force and the normal reaction can act through the rod’s centre of gravity (C)

That means there will be zero torque about the C and no rotation occurs.

(For this to work there is probably a requirement for the coefficient of friction to be less than some critical value.)

The original question is just a special case of this – with zero friction and α = 90º.

 

[alan123hk]

uppose there is some friction and the rod is ‘leaning backwards’ as it slides down (angle α between slope and rod). For some value of α, the resultant of the frictional force and the normal reaction can act through the rod’s centre of gravity (C) That means there will be zero torque about the C and no rotation occurs.

Is it like a skier coming down from a snowy mountain?

 

[Baluncore]

You could glue a big vertical rod to the top of a small horizontal toy car, (maybe Hot Wheels?) Then place it on a ramp, so the rod is perpendicular to the slope, and release it. If the wheels are small, and the friction low, it should roll down the slope without leaning forwards or backwards, until it reaches the end of the ramp.

 

[James Brown]

Ok there is a conclusion here. Actually, the rod will topple or not depends on how you place the rod. If you place the rod perpendicular to the slope, then yes, like @Baluncore said, the rod won't topple. Although it looks incredible, but if you think deeply it make sense. The reaction force that act on the rod can't make the rod spin as the net torque about the center of weight is zero. Some may said that there is net torque about the point of contact due the reason that weight is acting on it, but this is wrong. First of all, if the rod "wants" to spin, then what "helps" it to spin?Reaction force? Definitely not, it is pointed to the center of weight which means the net torque about the center of weight is zero mentioned before. If nothing helps the rod to spin, I'm sure that the rod will slide due the above reason. If you still don't understand, let us do a test: Remove that reaction force, the rod will accelerate without spinning. Now add that reaction force back, will this force cause the rod to spin? definitely not. If the rod is not perpendicular to the slope then it will be the opposite, the rod will rotate due to the reason that there is reaction force acting on it.
The mistake that I made was that I thought the weight will ONLY act on the center, yeah, this is a grave mistake. A simple experiment will prove that, use a scissor to cut a rubber band into a string, then drop it horizontally, you'll see that the shape of the rubber band won't change. If weight only act on the center of the rubber band then the shape will change because only the center will accelerate and other parts of the rubber band will also accelerate but not as fast as the center.

 

[Melbourne Guy]

Case 1(The force is acting perpendicular to the rod but not in the center part of the rod):

In this case, will the orientation of the rod change? Yes, because that point need to bring the whole rod to accelerate. It is much easier for that point of the rod to bring the left part of the rod to accelerate but it is harder to bring the right side of the rod to accelerate so the rod will spin.

What exactly is 'the force' here, @James Brown? I'm missing the context as it applies to the OP, because we know from free fall experiments, objects don't change orientation in the way I think you're describing.

 

[Steve4Physics]

Is it like a skier coming down from a snowy mountain?

I think skiers generally lean forwards when going downhill for reasons of stability/control/reducing air resistance.

You can't realLy compare a rod balanced on one tip against a skier with load distributed over the area of long skiis.

 

[jbriggs444]

But isn't every part of the rod being acted on by the same gravitational force, @alan123hk? Especially in your second diagram. If friction does not impart any force on the contact point, why would it move any faster or slower than the other end?

Because the normal force acts on the contact point alone.

 

[James Brown]

What exactly is 'the force' here, @James Brown? I'm missing the context as it applies to the OP, because we know from free fall experiments, objects don't change orientation in the way I think you're describing.

Ahh I take back all what I said about the rod, it does not contribute to the question

 

[jbriggs444]

I think skiers generally lean forwards when going downhill for reasons of stability/control/reducing air resistance.

If you fail to lean forward relative to gravity, your skis will slide out from under you and you'll land on your butt. Same would hold for ice skates. It's just that ice tends to be level.

 

[Melbourne Guy]

If you fail to lean forward relative to gravity, your skis will slide out from under you and you'll land on your butt.

Is that due to friction, @jbriggs444?

 

[jbriggs444]

Is that due to friction, @jbriggs444?

No.