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

In summary: Welcome to PF. In summary, the problem is that, someone told me that a ball won't roll when sliding down a frictionless slope because the resultant force mgsinx is parallel to the slope which means that the ball will slide down 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? Actually, I think the torque is not "around the point of contact" but rather around the center of mass of the rod.
  • #71
Melbourne Guy said:
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.
 
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  • #72
Baluncore said:
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.
 
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  • #73
DaveC426913 said:
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?
 
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  • #74
Baluncore said:
...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.
 
  • #75
DaveC426913 said:
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
Baluncore said:
Cancel all friction, ...
 
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  • #76
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. :smile:
 
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  • #77
Delta2 said:
...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?
 
  • #78
Melbourne Guy said:
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.
 
  • #79
Melbourne Guy said:
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.
 
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  • #80
jbriggs444 said:
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.
 
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  • #81
Melbourne Guy said:
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.
 
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  • #82
Delta2 said:
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?

Delta2 said:
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 👍
 
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  • #83
Melbourne Guy said:
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.
 
  • #84
Melbourne Guy said:
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.
 
  • #85
Lnewqban said:
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?
 
  • #86
Melbourne Guy said:
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.
 
  • #87
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.

1655018242669.png
 
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  • #88
alan123hk said:
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?
 
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  • #89
Melbourne Guy said:
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.. :smile:
 
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  • #90
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.
 
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  • #91
Steve4Physics said:
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?
 
  • #92
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.
 
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  • #93
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.
 
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  • #94
James Brown said:
Case 1(The force is acting perpendicular to the rod but not in the center part of the rod):
rod-experiment-1-png.png
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.
 
  • #95
alan123hk said:
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.
 
  • #96
Melbourne Guy said:
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.
 
  • #97
Melbourne Guy said:
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
 
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  • #98
Steve4Physics said:
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.
 
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  • #99
jbriggs444 said:
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?
 
  • #100
Melbourne Guy said:
Is that due to friction, @jbriggs444?
No.
 
  • #101
jbriggs444 said:
No.

It is due to the lack of friction.
Thanks, so does that mean the rod will topple? For the same reason? Or because the rod is more rigid than a person on skis, is the analogy not equivalent.
 
  • #102
Melbourne Guy said:
Thanks, so does that mean the rod will topple? For the same reason? Or because the rod is more rigid than a person on skis, is the analogy not equivalent.
I actually ran the experiment yesterday. Yes, the rod will topple.

A pencil-sized rod with a crudely exact hand-held position starting on end touching a handy sloping surface topples very quickly indeed. The direction of the topple for any given trial can nonetheless be seen. It depends on the initial lean angle. The neutral angle for my experimental setup was forward from the vertical and a bit backward from the perpendicular, exactly as one would expect.

One moment while I run a trial for a vertical pencil released on a 45 degree slope... It topples backward.

Try it. It is not hard to set up. All it takes is a book and a pencil.
 
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  • #103
jbriggs444 said:
I actually ran the experiment yesterday. Yes, the rod will topple.

A pencil-sized rod with a crudely exact hand-held position starting on end touching a handy sloping surface topples very quickly indeed. The direction of the topple for any given trial can nonetheless be seen. It depends on the initial lean angle. The neutral angle for my experimental setup was forward from the vertical and a bit backward from the perpendicular, exactly as one would expect.

One moment while I run a trial for a vertical pencil released on a 45 degree slope... It topples backward.

Try it. It is not hard to set up. All it takes is a book and a pencil.
Indeed it will... There is friction acting on it
 
  • #104
James Brown said:
Indeed it will... There is friction acting on it
Friction which would tend to make it topple forward. Yet it topples backward!
 
  • #105
jbriggs444 said:
I actually ran the experiment yesterday. Yes, the rod will topple.
With respect, @jbriggs444, the OP asked about a frictionless surface. Does a "crudely exact hand-held position starting on end touching a handy sloping surface" really constitute a meaningful investigation of the question?
 
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