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

[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 $F_n$

 

[alan123hk]

OK, here's what I can say.

​

But I still don't know $\theta~$, it's actually the value I'm looking for, so what should we do next?

Edit :
Another interesting thing is $~tan~\theta=\frac{F_n}{F_f}=\frac {1}{u}~~ \Rightarrow~~\theta={tan}^{-1}\left( \frac{1}{u}\right)$
If we substitute it into above equation, we can get $F_n$
But isn't $\theta$ the answer I am looking for? Why can it be found directly by $u$?
If so, why do I need to do so many other complicated derivations? I feel really lost.
Maybe it's as simple as that if friction doesn't vary with speed and no other complex stuff is involved.

 

[jbriggs444]

Another interesting thing is $\theta={tan}^{-1}\left( \frac{1}{u}\right)$

But isn't $\theta$ the answer I am looking for? Why can it be found directly by $u$?
If so, why do I need to do so many other complicated derivations? I feel really lost.

If you look up at #108, you may notice:

arc cotangent of the coefficient of friction

The free body diagram, writing down the force balance equations and torque balance equations and solving the resulting et of simultaneous equations is the brute force crank-and-grind method to solving problems of this sort. It essentially always works.

But more efficient shortcuts can exist. For the frictionless case, a number of participants here were able to immediately intuit that "perpendicular to the slope" was the answer. That answer also falls out of the arctangent of one over the coefficient of friction or [equivalently] the arc cotangent of the coefficient of friction

 

[alan123hk]

The free body diagram and writing down the force balance equations and torque balance equations and solving the set of simultaneous equations is the brute force crank-and-grind method to solving problems of this sort. It essentially always works. But more efficient shortcuts can exist. For the frictionless case, a number of participants here were able to immediately intuit that "perpendicular to the slope" was the answer. That answer also falls out of the arctangent of one over the coefficient of friction or [equivalently] the arc cotangent of the coefficient of friction

Okay, I'm basically satisfied with my understanding of the problem now. As for the advanced calculation method, I need time to study it. Thank you very much for your help.

 

[alan123hk]

I finally got it, it turns out that the explanation is very simple, no complicated math.

In the absence of friction, no matter what the shape of the object, as long as the slope normal vector at the point where the slope intersects the object passes through the object's center of mass, the normal force $F_n$ will not produce a twisting force and the object will remain unrotated. Conversely, if this normal vector does not pass through the object's center of mass, then a twisting force is created to rotate the object.

In addition, regardless of the rotating state or the non-rotating state, the acceleration of the object's center of mass sliding down the slope remains the same and should be equal to $\frac{F_g~sin(\alpha)}{m}$. But if the object is rotating, the torque applied to it seems to vary with its angle and profile, so its angular velocity change should be a fairly complex process.