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

[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.

 

[Melbourne Guy]

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.

 

[jbriggs444]

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.

 

[James Brown]

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

 

[jbriggs444]

Indeed it will... There is friction acting on it

Friction which would tend to make it topple forward. Yet it topples backward!

 

[Melbourne Guy]

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?

 

[jbriggs444]

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?

Yes, it does.

Forum rules and good manners suggest that I should leave it at that.

 

[alan123hk]

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.

Can I ask a question ?
Assuming that the rod has uniform density and mass $M$, and the friction coefficient is $c$, how can the normal direction reaction force, velocity, and the angle$~~\alpha~$ between the inclined plane and the rod be calculated to achieve this non-rotating state ?

 

[jbriggs444]

Can I ask a question ?
Assuming that the rod has uniform density and mass $M$, and the friction coefficient is $c$, how can the normal direction reaction force, velocity, and the angle$~~\alpha~$ between the inclined plane and the rod be calculated to achieve this non-rotating state ?

If all else fails, pick a coordinate system, generate your free body diagram, write down a force balance for the x and y coordinates along with a torque balance and solve. You'll want to have the angle of the plane as one of your inputs.

Edit to add...

If algebra is not your thing, one can short-cut the calculation by realizing that if the rod is sliding the ratio of the frictional force parallel to the slope and the normal force perpendicular to the slope is given by the coefficient of friction. So the neutral angle $\alpha$ measured as a a deviation from parallel to the slope is immediately given by the arc cotangent of the coefficient of friction. If the coefficient of friction is zero, the angle is 90 degrees and the rod is perpendicular to the slope.

If the rod is not sliding then things are even easier. The rod will balance when it is vertical.

 

[berkeman]

I need to ask for the Cliff Notes version of this thread. The Mentors are receiving complaints about it, and I need to figure out whether to tie it off and if so how.

Is the question about a frictionless plane or not? If there's friction, it seems like the tipping question is a standard problem of the FBD, etc. If it's frictionless, the question becomes more interesting, but has there been a solution that shows any relationship to the initial angle of the bar to the vertical in that situation?

Thank folks. Since I'm the newbie Mentor, the other Mentors are making me handle this one...

 

[hutchphd]

My name is not Cliff, but here is my version. This is not complicated IMHO

Absent friction there are two forces. One (call it $\vec N$) is normal to the ramp surface and passes up through the contact point. The other (call it $\vec W$) points straight down through the Center of Mass. Iff the line of action of $\vec N$ passes through the Center of Mass the object will not rotate.
For most objects this is possible but inherently not stable. But in the absence of perturbations, the CM will simply accelerate down parallel to the ramp according to $$m\vec a_{parallel}=\vec W sin (\alpha)$$ and no rotation will accrue if the initial geometry of the CM is correctly chosen.

 

[alan123hk]

Can I ask a question ?
Assuming that the rod has uniform density and mass M, and the friction coefficient is c, how can the normal direction reaction force, velocity, and the angle α between the inclined plane and the rod be calculated to achieve this non-rotating state ?

It seems likely that this thread will be mercilessly shut down, although if it does then the decision is certainly reasonable. I hope to get a chance to post before closing

I think this is just an ideal equilibrium state, which is actually very difficult to achieve, because acceleration is required before the steady state, and the friction force will increase with the speed change, which means that the angle of the long rod may keep changing during acceleration, it seems difficult to guarantee that it will eventually enter the non-rotating stable state smoothly.

 

[alan123hk]

If all else fails, pick a coordinate system, generate your free body diagram, write down a force balance for the x and y coordinates along with a torque balance and solve. You'll want to have the angle of the plane as one of your inputs.

This is my initial attempt, I'm checking it for errors, hopefully no big mistakes.

 

[James Brown]

I need to ask for the Cliff Notes version of this thread. The Mentors are receiving complaints about it, and I need to figure out whether to tie it off and if so how.

Is the question about a frictionless plane or not? If there's friction, it seems like the tipping question is a standard problem of the FBD, etc. If it's frictionless, the question becomes more interesting, but has there been a solution that shows any relationship to the initial angle of the bar to the vertical in that situation?

Thank folks. Since I'm the newbie Mentor, the other Mentors are making me handle this one...

This question is about a FRICTIONLESS plane

 

[jbriggs444]

This is my initial attempt, I'm checking it for errors, hopefully no big mistakes.

I do not understand the free body diagram in #112.

I see four forces depicted. There is $F_1$ acting one tip of the rod, $F_2$ acting at the other tip, $F_g$ acting at the same point as $F_1$ and $F_c$ acting at some point not even on the rod.

What are those four forces?
Which (if any) is the normal force of slope on rod?
If $F_g$ is gravity, why is it depicted acting at the endpoint of the rod?
Where is the slope?

 

[alan123hk]

What are those four forces?
Which (if any) is the normal force of slope on rod?
If Fg is gravity, why is it depicted acting at the endpoint of the rod?
Where is the slope?

$F_g~$ is the downward force on the rod due to gravity, it should actually be at the midpoint of the rod.
$Fc~$ is the frictional force acting on the lower end of the rod when it slide on the surface with slope $\alpha$
$F_1~$ is the torque applied to the upper end (or midpoint) of the rod, this force is perpendicular to the rod
$F_2~$ is the torque applied to the lower end of the rod, this force is perpendicular to the rod
The angle between the rod and the slope surface is $\theta$
The angle between the rod and the vertical downward gravity is $\beta$

I also feel like my description would be questioned, actually I'm not sure if this is correct.

 

[jbriggs444]

$F_1$ is the torque applied to the upper end of the rod, this force is perpendicular to the rod
$F_2$ is the torque applied to the lower end of the rod, this force is perpendicular to the rod

Torques are not forces. If there are only two forces acting on the rod then there should be only two forces shown on the free body diagram.

$F_c$ is the frictional force acting on the lower end of the rod when it slide on the surface with slope $\alpha$

Oh. The dotted line through $F_c$ is not the vector associated with $F_c$. It is, instead, a perpendicular to the torque $F_2$. And the slope coincides with the vector $F_c$ (which I'd have named $F_f$).

Artistically, I'd have drawn $F_c$ as a vector near by, parallel to and shorter than the slope and given it a weight or color different from that of the slope. But I am a terrible artist.

If you are going to label angle $\theta$, put the label on the angle between the rod and the slope, not on the angle between the rod and a dotted line with no well defined angle.

Torques do not have lines of action (even if torque $F_2$ actually existed). So the dotted line perpendicular to the torque $F_2$ is meaningless. In any case, for a stable equilibrium orientation, angular momentum about the rod's center of mass must be constant at zero. Accordingly the net torque about the rod's center of mass must be zero.

So let us make those modifications to your free body diagram: Remove $F_1$ and $F_2$. Move the line of action of $F_g$ to the midpoint of the rod and rename $F_c$ to $F_f$.

Now where is $F_n$?

We need to finish the free body diagram correctly before we can move on to writing down the force and torque balances.

 

[alan123hk]

So let us make those modifications to your free body diagram. Remove F1 and F2. Move the line of action of Fg to the midpoint of the rod.
Now where is Fn?

These two torques are the basis of my inference, and removing them is actually a complete negation of my entire inference.

Now where is Fn?

What does $F_n$ mean?
Is it a normal force exerted on something?

 

[jbriggs444]

What does $F_n$ mean?
Is it a normal force exerted on something?

Yes. The contact force between rod and slope can be decomposed into two components:

The force of friction (parallel to the slope)
The normal force (perpendicular to the slope).

The normal force will be whatever it has to be to prevent the rod from crunching through the slope.

 

[alan123hk]

Yes. The contact force between rod and slope can be decomposed into two components:
The force of friction (parallel to the slope)
The normal force (perpendicular to the slope).

If I'm not mistaken, I think what you mean is as shown below.

​

I don't know how to get $~F_n~$ because I don't know $~F_c~$ yet, even if we assume $~F_f=uF_n~$ where $u$ = Sliding Friction. Should I find $~F_c~$ first and how? Could you please give further hints ?

 

[hutchphd]

That looks good. I would just assume the usual $$F_f=\mu F_n$$ and let it be the same for static or sliding.
Write down Newton's equations for directions $\perp$ and for $\parallel$ to the plane. Then$$a_\perp=0$$ will tell you $F_n$