What "really" produces motion on incline plane?
Author
EDIT:
Just for the record this is not homework. I would be happy if my high school taught this stuff. I am tryign to understand stuff that a physics book I am reading doesn't specifically mention.
Okay, say we have got a cylinder lying down on an incline plane. You are looking at it from the side so you just see a circle. There is a coefficient of friction u between the cylinder's surface and the plane's surface. The angle of the plane is theta, and the cylinder has some radius r.
I understand that the force down the plane is:
mgsin(theta)
I understand that the friction force is:
u*mgcos(theta), which says the normal force * friction coefficient
That was the relatively easy stuff. The center of gravity is just the center of the cylinder, the inertia tensor is just a scalar in this case because it is a cylinder on a single rotation axis, and it has some funky value (specifically what it is I dont' care right now).
The cylinder will do two things: move down the plane linearlly and start rotating.
Now, my first question: What is the force that produces the torque? Is it the friction force which produces torque? Or is it just the normal force? They are almost the same, the normal force is mgcos, while the friction force is u*mgcos. They both produce the same axis of rotation too. I am thinking it is just the friction force.
Second, what happens when the angle of theta is such that the friction force cancels the force down the incline plane? Will the object still start to move because of the normal force which generates a torque, which then in turn pushes off of the surface down the incline plane because of friction?
Third, when the angle theta exceeds a given value, is it true that a) the cylinder will keep sliding down the plane but b) the angular acceleration stays at whatever it was for the maximum value of the friction force up the plane.
I'm having a difficult time formulating in words exactly what I'm asking and I can try to refine them if they do not make sense.
[edited by - shadow12345 on May 9, 2004 10:12:03 AM]
[edited by - shadow12345 on May 9, 2004 10:13:58 AM]
[edited by - shadow12345 on May 9, 2004 10:14:47 AM]
[edited by - shadow12345 on May 9, 2004 10:16:00 AM]
1. As long as the cylinder doesn''t slide down the plane, the only non-moving point at any moment is the point of contact, which is the point in relation to which you have to compute the torque of all forces. The normal force and friction force being applied on that point, they have no torque as far as that point is concerned. However, the weight (gravitational force) being applied to the object produces a torque which gets it moving (the cylinder rolls down because its center of gravity is not above the point of contact).
Torque must be calculated with respect to a point whose velocity is zero (the instantaneous axis of rotation).
2. If the friction force is strong enought, we can assume we''re in the above case (the cylinder doesn''t slide), so only the gravitational pull will act, and the cylinder will roll down the plane.
3. The force of static friction (which prevends sliding) is not the same as that of "movement" friction. As long as the point of contact has a null velocity, only the static friction forces apply. Once, however, these forces are not enough to keep the point of contact at a null velocity, they break and aren''t applicable anymore : now, the object is sliding, so another kind of force appears (which you have to study). The static friction forces will only set back in once the relative speed at the point of contact is zero again.
Any (more) questions?
Torque must be calculated with respect to a point whose velocity is zero (the instantaneous axis of rotation).
2. If the friction force is strong enought, we can assume we''re in the above case (the cylinder doesn''t slide), so only the gravitational pull will act, and the cylinder will roll down the plane.
3. The force of static friction (which prevends sliding) is not the same as that of "movement" friction. As long as the point of contact has a null velocity, only the static friction forces apply. Once, however, these forces are not enough to keep the point of contact at a null velocity, they break and aren''t applicable anymore : now, the object is sliding, so another kind of force appears (which you have to study). The static friction forces will only set back in once the relative speed at the point of contact is zero again.
Any (more) questions?
Victor Nicollet, INT13 game programmer 
Author
Hi, I just wanted to let you know that a while back you wrote a 'matrix primer' on these threads and I copied that to a text file, put your name at the top and your email address so I remembered who wrote it, and I still use it and was wondering if I could keep it in the folder when I send it (my graphics programming projects) to friends and stuff.
I had a question about this:
I am confused about this. It seems that something else must be involved, i.e somehow the normal force or friction force must be responsible for the rotation.
EDIT:
like, what I mean is, mg pulls down, but then there is a point of contact which is not inline with the center of gravity (like you said cg hangs out in front of point of contact). Then, the normal force which pushes back upwards must be responsible for the torque (but it also seems like it could be friction force, but it cannot be both).
Also, are the coefficients of static and kinetic friction calculated experimentally only, or can it be done by examining chemical properties somehow?
[edited by - shadow12345 on May 9, 2004 10:38:13 AM]
[edited by - shadow12345 on May 9, 2004 10:39:04 AM]
[edited by - shadow12345 on May 9, 2004 10:40:43 AM]
I had a question about this:
quote:
(the cylinder rolls down because its center of gravity is not above the point of contact).
I am confused about this. It seems that something else must be involved, i.e somehow the normal force or friction force must be responsible for the rotation.
EDIT:
like, what I mean is, mg pulls down, but then there is a point of contact which is not inline with the center of gravity (like you said cg hangs out in front of point of contact). Then, the normal force which pushes back upwards must be responsible for the torque (but it also seems like it could be friction force, but it cannot be both).
Also, are the coefficients of static and kinetic friction calculated experimentally only, or can it be done by examining chemical properties somehow?
[edited by - shadow12345 on May 9, 2004 10:38:13 AM]
[edited by - shadow12345 on May 9, 2004 10:39:04 AM]
[edited by - shadow12345 on May 9, 2004 10:40:43 AM]
Concerning the matrix primer, I''d be interested to know how far it goes 
in that case, you''d better add my new e-mail address to it (victor.nicollet@free.fr).
What you have to ask yourself is : what is the cylinder rotating around? Sure, if you (as an observer) are moving along with the cylinder, then the center of rotation will also be the cylinder''s center -- but in that case, things won''t be gallilean anymore and you have to take into account inertial forces as well.
But an observer that''s sitting on the plane will see that all the points of the cylinder are moving, except for the point of contact. While this is difficult to see on a cylinder, it''s quite obvious if you replace it by a cube and flip it over (it will rotate around one of its corners). So the real axis of rotation, in the gallilean referential of the laboratory, is the line of contact between cylinder and plane.
And the only force which has a torque with respect to this axis is the gravitational pull.
OK, I''ve already said this above. So, what about the friction force''s addition to the rotation? Well, it''s still here, but it''s hidden in the "no sliding" hypothesis. If the point of contact has no velocity, it''s because the friction force prevents it from moving away. If it wasn''t for friction, then, the rest of the cylinder would simply slide down the plane without any rotation.
So I guess my "rolls because of gravity" sentence was somehow wrong because it didn''t take into account that no-sliding hypothesis. What I should have said is either "moves down because of gravity", or "rolls because the contact point is held still and gravity makes it tip over".
in that case, you''d better add my new e-mail address to it (victor.nicollet@free.fr).What you have to ask yourself is : what is the cylinder rotating around? Sure, if you (as an observer) are moving along with the cylinder, then the center of rotation will also be the cylinder''s center -- but in that case, things won''t be gallilean anymore and you have to take into account inertial forces as well.
But an observer that''s sitting on the plane will see that all the points of the cylinder are moving, except for the point of contact. While this is difficult to see on a cylinder, it''s quite obvious if you replace it by a cube and flip it over (it will rotate around one of its corners). So the real axis of rotation, in the gallilean referential of the laboratory, is the line of contact between cylinder and plane.
And the only force which has a torque with respect to this axis is the gravitational pull.
OK, I''ve already said this above. So, what about the friction force''s addition to the rotation? Well, it''s still here, but it''s hidden in the "no sliding" hypothesis. If the point of contact has no velocity, it''s because the friction force prevents it from moving away. If it wasn''t for friction, then, the rest of the cylinder would simply slide down the plane without any rotation.
So I guess my "rolls because of gravity" sentence was somehow wrong because it didn''t take into account that no-sliding hypothesis. What I should have said is either "moves down because of gravity", or "rolls because the contact point is held still and gravity makes it tip over".
Victor Nicollet, INT13 game programmer
A nice interesting problem, pretty easy considering it''s worth 20 marks on a past paper of mine. Here goes:
A cylinder is placed on an inclined frictionless plane. The cylinder just slides down. Now the plane is roughened, so instead the cylinder rolls down. Show that now it takes sqrt(3/2) times the time to roll down the slope.
Nice and easy, I think it was worth loads of marks because it was expected to be rigorous, ie calculating the moment of inertia of a cylinder etc.
A cylinder is placed on an inclined frictionless plane. The cylinder just slides down. Now the plane is roughened, so instead the cylinder rolls down. Show that now it takes sqrt(3/2) times the time to roll down the slope.
Nice and easy, I think it was worth loads of marks because it was expected to be rigorous, ie calculating the moment of inertia of a cylinder etc.
Author
quote:
And the only force which has a torque with respect to this axis is the gravitational pull.
Okay, I see exactly what you are saying, and I am now thinking of the problem in a different manner. There is a ''line'' on the cylinder that touches the plane which the cylinder rotates about. This line of contact does not move, and the friction holds it there in place. If there was no friction, this ''line'' would not be held in place, and the cylinder would slip and not rotate. The torque is then the force downward due to gravity, times the distances that the center of gravity hangs out from this line of contact.
Now, does the actual *value* of friction affect how the cylinder will rotate? I.e, can you compute some angle of theta, with a given value of coefficient of friction, in which the cylinder will start to slip? I.e, I took a notebook and put a pen on it. For small angles of incline it just roles down with no slipping. Once you get to a certain angle however, it seems that the pen moves linearly down the notebook and rotates, but that the distances it moves linearlly is *greater* than the linear distance of a point on the circle which represents the cylinder. I remember the linear distance of a point on a circle being the degrees in radians * radius.
There are two ways to look at this problem. One way is to look at it from the point of view of the contact point between the cylinder and the plane, which is what ToohrVyk was talking about. The other way to look at it is from the point of view of the cylinder''s center. The cylinder''s weight acts on the center of mass so it has no torque. The normal force points from the contact point towards the center of mass, so it is parallel to the vector joining the center of mass to the contact point, and it thus also has no torque (think about it - you can''t rotate something by pressing on it towards its center, you would just translate it). The frictional force is the one that puts the torque, equal to μ mg cos θ r, so the angular acceleration is α = τ / I = (μ mg cos θ r) / I. The total sum of the forces parallel to the plane is not simply mg sin θ - the friction points up the plane, so the net force down the plane is mg sin θ - μ mg cos θ, so the linear acceleration of the cylinder is (mg sin θ - μ mg cos θ ) / m = g sin θ - μ g sin θ. Note that even if static friction cancels the net force down the plane, there is still a net torque, so the cylinder is not in equilibrium. Once it starts rotating, kinetic friction will kick in, so then the cylinder will start to slide down the plane. It''s difficult to describe.
Author
No, you did a very good job of describing that. I think that I don''t have anymore questions you both did very good jobs explaining this. Thanks!
About the matrix primer, I will put the new email address in there and I can tell you about any responses to it or something like that
Thanks again guys.
About the matrix primer, I will put the new email address in there and I can tell you about any responses to it or something like that
Thanks again guys.
Let me clarify what happens when the frictional force cancels the component of the weight down the plane. There are two cases - static friction and kinetic friction. If you have static friction, then the frictional force is only as large as it needs to be to keep the contact point stationary. It would be the value that makes it so that a = αr, where a is the linear acceleration of the cylinder and α is the angular acceleration. When this happens, the static friction is less than μsmg cos θ, so it doesn''t actually cancel the planar component of weight and the cylinder rolls down the ramp. In the case of kinetic friction, when μkmg cos θ equals mg sin θ, then the cylinder will slide down the ramp at constant velocity, but its angular velocity would increase (kind of bizarre, isn''t it?) until static friction acts once again, and the sliding stops and rolling continues.
Author
quote:
(mg sin ? - ? mg cos ? ) / m = g sin ? - ? g sin ?
did you mean it equals g(sin - ucos) instead of g(sin - usin)
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