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

[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.

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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.