Pure rotation under a point force and a distributed friction

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1. May 6, 2016

onder

Hi guys,

I was wondering if it is possible to have a pure planar rotation of a rectangular-prism shaped rigid body on a planar surface when it is subjected to a planar point force at the tip. Is there any range for the point force such that it can not break the static friction force (so no acceleration) but breaks the static friction torque around the center of mass (COM) of the body and thus, causes a pure rotational motion? I appreciate your answers with at least semi-theoretical proofs.

Thanks

2. May 6, 2016

haruspex

The static friction does not understand the distinction you are making. At each point of the contact area, it is just force opposing tendency to relative motion.
Pick a point P as the potential axis of rotation. When about to rotate around that point, for any small surface element A in the contact area, the frictional force will be normal to PA and its magnitude will be the product of the coefficient of static friction and the weight above A. That force will have a moment about P opposing the moment of the applied force.
When it slips, it will be about that P for which equality between applied moment and available static friction moment is first reached.
That might make it sound all too hard, but consider how this compares with inertia, without friction. The linear acceleration of an element of mass will be proportional to its distance from the axis of rotation. So the inertial resistance behaves the same way as frictional resistance, and the initial point of rotation will be the same.

3. May 8, 2016

onder

Thank you! Could we say this point P is actually the center of mass of the object or could it be something else? It is a pretty easy question when there is a hinge or a fixed point. But it was a little complicated to me once I start to think this issue for a free body on a surface :)

4. May 8, 2016

haruspex

I should be honest and say I am not completely convinced by the argument in my previous post, but I don't see a flaw in it. anyway, assuming it is right, we can treat it as an impact question.:
A rectangular block mass m lies on a surface. It is struck on the side at one end at right angles with momentum p. Where is the instantaneous centre of rotation?
The height doesn't matter. Say it has length 2x on the side struck and 2y on the other side (the direction of the blow).
By symmetry, the centre of rotation will be half way across from the struck side (y). Suppose it is r from the mass centre in the x direction.
By linear momentum, the mass centre moves off at speed v=p/m. Taking moments about the mass centre, conservation of angular momentum gives Iω=px. We also have v=ωr. Combining, mrx=I.
For a rectangular block moment of inertia about its mass centre is m(x2+y2)/3, so $r=\frac{x^2+y^2}{3x}$.

5. May 10, 2016

onder

Thanks a lot for your answer! I have 1 objection to this, we are assuming the center of mass would go with the velocity v=p/m and also we say there is a rotational speed ω around $r=\frac{x^2+y^2}{3x}$ . In this case, due the rotation of the robot, which is not at the center of mass, the center of mass has additional linear velocity, let's say $v_r = d \times ω$, where d is the distance vector from center of mass to the center of rotation point. If I am right, that should be against the conservation of momentum because that results in increase (or decrease) in the linear momentum of the overall system.

6. May 10, 2016

haruspex

I think you must have misunderstood my reasoning.
We can think of the motion as a linear motion of the mass centre plus a rotation about the mass centre.
I set the linear speed of the mass centre according to the conservation of momentum, and the angular speed from conservation of angular momentum. At the instantaneous centre of rotation, these cancel to produce (momentarily) no net velocity.

7. May 10, 2016

onder

Ah, now it is clear!! I was looking from the object's coordinate frame! So basically, the point r at your derivation is the location where |v|=0 at the global frame. Would the following statement be true then: could we say the center of rotation coincides with the center of mass if the global coordinate frame moves together with the center of mass? The second question is, if there is a distributed resistive force such as friction, would the location of the center of rotation be different?

8. May 10, 2016

Nidum

Analysis of this problem requires consideration of :

(1) The redistribution of weight load and consequent effect on friction forces acting in different regions of the contact area which occur as a result of the line of action of the applied force being a vertical distance above the prism base .

(2) The effect of imperfections . Some of which are :

Variations of coefficient of friction and related friction force in different zones in the contact area .
Variations of direction and offset of the line of action of the applied force .

Very small imperfections could alter the response motion of the prism considerably .

Analysis of this problem is essentially the same as analysis of stability of sliding bearings and brake pads .

Last edited: May 10, 2016
9. May 10, 2016

haruspex

Having thought through it some more, my argument relating to angular momentum doesn't work. That has an additional factor of distance, making it mr2 (of course), not mr. It might be hard to figure out the initial centre of rotation, even for a simple rectangle, and making the simplifying assumption of uniformly distributed normal force. On top of that are the complexities Nidum mentions.

10. May 11, 2016

haruspex

Let's simplify it a little more: a thin uniform bar length 2L mass m lies on a horizontal surface with coefficient μk. It is pushed with a perpendicular horizontal force at one end. It rotates about a point x from its mass centre at constant speed.
As it turns through angle θ, find the work done by the force and the work done against friction. Equate the two. Find the x which minimises F.

11. May 13, 2016

onder

Thanks, guys! I like simplifying, so let's focus on this simplified case. I am just thinking in terms of "first motion" which is when it just starts to rotate. It is still complicated to me because I am not sure what the friction distribution should be. Once we apply the force, there should also be a resistive torque due to the torque contribution of each friction elements. Could you hint more how to find x? Thanks!

12. May 13, 2016

haruspex

Assuming the normal force is uniformly distributed along the bar, the frictional force will be too. If the bar has linear density ρ, what is the mass of an element length ds? What is the frictional force on it when about to slide? In what direction is that force?
If the element is at distance s from the point X, what is the moment of it about X?

13. May 16, 2016

onder

Exactly, so the question is, where should be the pivoting point for the rotation?

14. May 17, 2016

haruspex

I gave you the start of the way to find that. You can answer the question I asked in post #12 in terms of the unknown distance s. Later on we can get an equation that you can solve to find s.

15. May 17, 2016

onder

Thanks! I have already implemented a numerical approach to solve that and found the frictional force distribution on infinitesimal elements. I just don't know where the pivotting point is. Could you define what is the pivotting point in this case?

16. May 17, 2016

haruspex

It will be that point where the ratio of the moment of the applied force to the moment of the static frictional force is greatest. That is why the first step is to say let the point be distance s along the rod from the applied force. What is the moment of the applied force about the point?

17. May 17, 2016

onder

Thanks a lot, haruspex!! That was a great thread for me. For the defined s above, the moment of the applied force is going to be (-s x F). I will figure out where the s is. It seems like it will be the CM for the whole motion but I am not sure. At least for the first motion, it should be CM because the moment of static friction on the CM will be 0 where the ratio is maximized.

"It will be that point where the ratio of the moment of the applied force to the moment of the static frictional force is greatest."
That sounds right to me but I can not fully convince myself. Would it be possible to have at least a semi-proof for this? Also, the friction is specified as static friction which is probably the consideration of the "first motion". In addition to that, I was wondering would this statement still be valid if the object is already moving?

18. May 17, 2016

haruspex

For the semi-proof (or even, proof?), consider increasing the applied force gradually until its moment about some point matches the maximum moment of the static frictional force about the same point. The point it will rotate about will be that point for which the applied force reaches that level first, i.e. the one for which the ratio is greatest.
This will work out the same for both kinetic friction and static friction. The complication is that assuming kinetic is lower then as soon as it starts to rotate it will accelerate (for the same applied force). Now the moment of inertia comes into play. The MoI of a mass element goes like r2, whereas the frictional moment goes like r. So the centre of rotation will shift.
(Doubt: am I saying that to find when it will start to move we have to use the frictional moment, but to find how it will move we use the sum of the frictional moment and the inertial moment? Ouch!)
To find s, you need to find the opposing frictional moment about that point. This will involve an integral. Consider an element length dx at distance x from the presumed rotation point. What is the frictional force on it, and what is its moment about the point?

19. May 20, 2016

onder

Now it makes a lot of sense to me (especially for the "first motion, static"). Once the object starts moving, now, depending on the velocity vector of each infinitesimal elements, the friction force is going to be the kinetic friction force acting towards the opposite direction of the velocity vector of the infinitesimal element.
Basically, at each instant during the motion, we know the force distribution along the body (which is applied force and kinetic frictional force). Now, we need to find the center of rotation point s. We already know the corresponding moments on this point because we know the force distribution. The corresponding moment of all the forces acting on the dynamic body (friction and actuation force) should be the integral of the moment of each infinitesimal forces which is $\int x X F \, dx$.
To summarize, there are 3 terms to be clarified to conclude the angular acceleration around the rotation point s:
1. moment around s due to friction, $M_f$,
2. moment around s due to the applied force, $M_{app}$,
3. and MoI around s, $I_s$

For a known s, it is easy to compute all these terms by integrating along x as you have indicated but what should be the s though? Or would it be possible for you to define the s under the dynamic motion of the free body? It was a very informative thread for me so I greatly appreciate for that. Once you could provide me the way how to find the point s for the dynamic case, the train of thought would be completed for me. Thanks!

20. May 21, 2016

haruspex

Before worrying about the accelerating case, let's try to solve the constant rotation case. Here we are assuming the system has started to move, by whatever means, but the applied force is only just sufficient now to match the kinetic friction. This means we can ignore moment of inertia and solve for balance of torques.
What is X there? Just multiplication?
What is F exactly? If it is a force then the integral does not make sense dimensionally. Remember we are looking at an element length dx of the rod. What is the normal force, so what is the kinetic frictional force?
When you think you have the right form of the integral, fill in the bounds and perform the integral.