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

• B
James Brown
Well, 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?

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• Delta2

Gold Member
...a ball won't roll when sliding downslope...
Not true. A simple test will bear that out.

And maybe don't trust that person to tell you things.

• vanhees71
James Brown
Not true. A simple test will bear that out.

And maybe don't trust that person to tell you things.
Well it is a frictioness slope, u mean the ball with also roll when sliding downa frictioness slope? Well if I trust him I won't be asking this question

• vanhees71
Gold Member
I missed the "frictionless" reference. mea culpa.

• berkeman, vanhees71 and hutchphd
Welcome to PF.
Well it is a frictioness slope, u mean the ball with also roll when sliding downa frictioness slope?
Without friction there will be no rotation, so the ball is in effect falling.
But there is no such thing as a real frictionless surface.

The rod will not topple in the frictionless model, but it will topple in practice.

• vanhees71 and Delta2
Gold Member
Without friction there will be no rotation, so the ball is in effect falling.
So, does that hold for the round-ended rod as well?

Now, replace the ball with a round headed rod, ...
Does it have a flat foot?
It matters not, head or foot, flat or round, the rod would slide down a virtual frictionless surface.

• Delta2
Gold Member
2022 Award
So, does that hold for the round-ended rod as well?
I am doubtful that the rod will stay upright. The momentum of the center of mass will, it seems to me, not allow the middle of the rod to keep up w/ the rounded bottom as it slides down so the rounded bottom will slide out from under the center of mass the the rod will topple.

@Baluncore I take it you do not agree w/ this.

... so the rounded bottom will slide out from under the center of mass ...
Then in your analysis, that must also be true of the ball.

Do you assume the rod is vertical or perpendicular to the surface?
... , does this means that the rod won't topple no matter the rod is perpendicular to the slope or not?

Gold Member
2022 Award
Then in your analysis, that must also be true of the ball.
No, because the normal force of the ball pressing against the slope runs THROUGH the center of mass so absent any impediment (friction) to the point of contract, the ball has no reason to roll. With the vertical rod, the normal force against the point of contact with the slope does NOT go through the center of mass, so there is an imbalance and the rod topples.

• Steve4Physics
With the vertical rod, ...
Ah. You assume the rod is vertical.
The OP assumed the rod could be perpendicular to the surface.

Homework Helper
Gold Member
@Baluncore what do you mean rod vertical and rod perpendicular, vertical and perpendicular are synonyms depending on the context. A scheme would help here.

If the rod is going to topple there must be a net torque around some point, I don't see any net torques here.

Homework Helper
Gold Member
Actually I think I am wrong, the weight has a net torque around the point of contact

Gold Member
2022 Award
Actually I think I am wrong, the weight has a net torque around the point of contact
Exactly. The rod does, the ball doesn't. See post #10

Actually, I think the torque is not "around the point of contact" but rather around the center of mass of the rod.

• Delta2
Homework Helper
Gold Member
I think both the ball and the rod have zero weight torque around their CM.

James Brown
Does it have a flat foot?
It matters not, head or foot, flat or round, the rod would slide down a virtual frictionless surface.
Omg this is incredible! Now do it in practice, place the rod on the slope, the rod will rotate without sliding at first right? Now, assume the friction is gone, if it is from your assumption then there is net torque so u think the rod will slide down without topple?

Homework Helper
Gold Member
We need an expert in rigid body dynamics here. I am a mathematician, my physics/mechanical intuition says that the rod will topple or not depending on the height of the rod , the diameter of the rounded head and the position of its CM.

• James Brown
James Brown
We need an expert in rigid body dynamics here. I am a mathematician, my physics/mechanical intuition says that the rod will topple or not depending on the height of the rod , the diameter of the rounded head and the position of its CM.
Yeah you are right

• Delta2
Homework Helper
Gold Member
Actually I keep switching between yes and no the more I keep thinking about this problem. My thoughts on this are a pure confusio so I just can't post them because even if I did no one, maybe not even me would understand what I said.

• James Brown
alan123hk
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?

I think that even if there is friction between the object and the slope, as long as the friction between all parts of the object in contact with the slope is the same, the object will not necessarily tip over in this equilibrium state.

• James Brown
what do you mean rod vertical and rod perpendicular, vertical and perpendicular are synonyms depending on the context.
The two special cases are with (1) the rod perpendicular to the slope, and (2) the rod vertical.
1. The rod would remain perpendicular to the slope as it slides, with all parts accelerating at the same rate. A sphere is an example of a short rod, with CofM on the perpendicular to the contact point on the slope.
2. A vertical rod would rotate about it's CofG, and would fall as the foot slid down-slope.

• Steve4Physics, hutchphd and Ibix
Homework Helper
Gold Member
I think that even if there is friction between the object and the slope, as long as the friction between all parts of the object in contact with the slope is the same, the object will not necessarily tip over in this equilibrium state.
I think we mean a perfect rounded head that has only one point of contact with the slope.

• alan123hk
I think we mean a perfect rounded head that has only one point of contact with the slope.
I think you mean a perfect rounded foot.

• • hutchphd and alan123hk
James Brown
Actually I keep switching between yes and no the more I keep thinking about this problem. My thoughts on this are a pure confusio so I just can't post them because even if I did no one, maybe not even me would understand what I said.
Same here, I think if the rod will topple or bot depends on how you put the rod

• Delta2
James Brown
I think we mean a perfect rounded head that has only one point of contact with the slope.
Indeed

2022 Award
I think this is the point @Baluncore made in #21 in diagrammatic form. There are only two forces here in the absence of friction - the green weight and the red normal reaction force. The weight has no moment about the center of mass of the rod so is irrelevant to whether or not it rotates. The normal reaction force, however, can cause it to rotate except in the special case that it points at the center of mass - i.e., when the rod is perpendicular to the slope. In the case shown, the rod will rotate clockwise as it slides down to the right, so "fall backwards".

In reality, a frictional force pointing up the slope will tend to cause the rod to rotate anticlockwise. Which way the rod actually falls depends on the balance between the torques from the normal and frictional forces.

• Delta2
James Brown
I think this is the point @Baluncore made in #21 in diagrammatic form.
View attachment 302676
There are only two forces here in the absence of friction - the green weight and the red normal reaction force. The weight has no moment about the center of mass of the rod so is irrelevant to whether or not it rotates. The normal reaction force, however, can cause it to rotate except in the special case that it points at the center of mass - i.e., when the rod is perpendicular to the slope. In the case shown, the rod will rotate clockwise as it slides down to the right, so "fall backwards".

In reality, a frictional force pointing up the slope will tend to cause the rod to rotate anticlockwise. Which way the rod actually falls depends on the balance between the torques from the normal and frictional forces.
Well so if the Reaction force does not point to the center of weight the rod will topple?

2022 Award
Well so if the Reaction force does not point to the center of weight the rod will topple?
Looks like it to me. If it feels slightly counter intuitive, I think it's the lack of friction messing with your notion of how stuff behaves.

James Brown
I think this is the point @Baluncore made in #21 in diagrammatic form.
View attachment 302676
There are only two forces here in the absence of friction - the green weight and the red normal reaction force. The weight has no moment about the center of mass of the rod so is irrelevant to whether or not it rotates. The normal reaction force, however, can cause it to rotate except in the special case that it points at the center of mass - i.e., when the rod is perpendicular to the slope. In the case shown, the rod will rotate clockwise as it slides down to the right, so "fall backwards".

In reality, a frictional force pointing up the slope will tend to cause the rod to rotate anticlockwise. Which way the rod actually falls depends on the balance between the torques from the normal and frictional forces.
Wait… the resultant force that act on a rod will becomes mgsinx right?

Homework Helper
Gold Member
@Ibix suppose the starting position for the rod is such that the normal force passes through the center of mass. What about the torque of mgsinθ (that is the component of weight parallel to the slope), wouldn't that make the rod topple? Why do we consider torque only around center of mass?

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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.

• DaveC426913
2022 Award
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.

• Delta2
Homework Helper
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.

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• Steve4Physics, alan123hk and Ibix