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Direction of friction when rolling without slipping 
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#1
Oct2610, 12:50 PM

P: 12

We've just studied rolling without slipping and I was having some trouble deciding the direction of the frictional force in some cases. So I pondered a little bit and recalled how you choose the direction of friction for "normal" translation without rotation:
Hope I didn't forget anything. Anyway, when rolling without slipping, the friction is static, so I presumed this static friction should have the same rules for deciding its direction than static friction in "normal" translation. And this worked! When I solved some exercises with this method, I got the right answer. This even explains that weird fact that when rolling downhill, friction is uphill, but when rolling uphill, friction is also uphill. This is because whether the body is rolling uphill or downhill, the only resultant force parallel to the surface is Mgsin([tex]\theta[/tex]), [tex]\theta[/tex] the angle of the incline, and this is always directed downhill, so the opposing static friction would always be uphill. Problem solved, right? Then, my professor gives us a test with a question about a yoyo that rolls without slipping, which is having its string pulled by someone. The question came with the following freebody diagram: Which doesn't make sense! The tension force exerted by the string is to the left; the friction should be to the right! Argh! So HOW exactly do you predict the direction of friction when rolling without slipping? Thanks. 


#2
Oct2610, 01:11 PM

P: 337

Let's say it isn't a yoyo but a bicycle wheel suspended in the air instead? Now as the string on the top of the bicycle wheel is being pulled what does the bottom of the wheel want to do? So in what direction would friction act if it tended to stop this action on the part of the bottom of the yoyo? 


#3
Oct2610, 02:49 PM

HW Helper
P: 7,033

Note for a hollow cylinder, the friction force would be zero. I assume the diagram is of a solid uniform cylinder. In the diagram, the string is generating a torque in addition to the surface the pulley rolls on, and it turns out that the torques are opposing, so the friction force is in the same direction as the string, but smaller. Comparing the rolling case to a frictionless surface case, in the rolling case, the solid cylinder will have more linear acceleration and less angular acceleration, since friction force is in the same direction as string tension. It might be easier to visualize this by inverting the situation and observing the speed of the upper surface of an object rolling and accelerating on an accelerating treadmill (the treadmill would be the string). The upper surface of the object moves backwards with respect to the ground, (except for a hollow cylinder where the upper surface doesn't move). Now imagine placing a board on top of the rolling object with it's backwards moving upper surface. It's rate of linear acceleration will increase, and it's rate of angular acceleration will decrease. That board is the equivalent of the friction surface on the bottom of the diagram above, and it's friction force would be in the same direction at the treadmill (string). Here is an example with real numbers, assume 1 newton of tension in the string, and a 1 kg uniform solid cylinder. frictionless case (+1 newton string tension, 0 friction force) cylinder linear acceleration = 1 m / s^{2} cylinder anglar acceleration / radians = (1 N r) / (1/2 kg r^{2}) = (2 m) / (r s^{2}) cylinder angular surface acceleration relative to cylinder = 2 m / s^{2} cylinder bottom surface accleration relative to plane = 1 m / s^{2} string acceleration relative to plane = 3 m / s^{2} rolling case (+1 netwon string tension, f = friction force in newtons) cylinder linear acceleration = (1 + f) m / s^{2} based on rolling, cylinder angular acceleration / radians = (1 + f) m / (r s^{2}) based on torques, cylinder angular acceleration / radians = ((1  f) N r) / (1/2 kg r^{2}) = 2 (1  f) m / (r s^{2}) (1 + f) m / (r s^{2}) = 2 (1  f) m / (r s^{2}) 1 + f = 2  2f 3 f = 1 f = 1/3 cylinder linear acceleration = 4/3 m / s^{2} cylinder angular acceleration / radians = 4/3 m / (r s^{2}) cylinder angular surface acceleration relative to cylinder = 4/3 m / s^{2} cylinder bottom surface accleration relative to plane = 0 m / s^{2} string acceleration relative to plane = 8/3 m / s^{2} The inclined plane case math for cylinders and spheres was done here: http://www.physicsforums.com/showthread.php?t=328621 


#4
Oct2810, 06:20 PM

P: 12

Direction of friction when rolling without slipping
Hmm, from what you guys said, I tried to approach the problem from a different angle...can some somebody check my work please?
I tried using LaTeX, but there's some weird error where I'm only getting thetas no matter what I type, so sorry for readability. There's a frictional force f = f i on the cylinder (yeah, the exercise said to treat the yoyo as a cylinder). I don't know the direction of this force; it will depend on the sign of f. The displacement vector of f with respect to the center of mass is r_{f} = R j, R the radius of the cylinder. Thus the torque caused by the frictional force is t_{f} = Rf (j x i) = Rf k. There's also the tension force T = T i due to the string. I'm going to say that this force is being applied from a distance of d*R from the center of mass, d between 0 and 1. If d = 0, the string is being pulled from the center of mass; if d = 1/2, halfway between the CM and the top of the cylinder; if d = 1, from the top, etc (it's a magical intangible string ). So the displacement vector is r_{T} = dR j. The torque is t_{T} = dRT (j x i) = dRT k. The resultant torque is t = (Rf  dRT) k. Because t = Iɑ and I = (1/2)MR^{2}, we get (Rf  dRT) k = (1/2)MR^{2}ɑ k (first equation). The resultant force is F = (T + f)i and from F = MA, we get our second equation: (T + f)i = MA i. Finally, the tangential acceleration due to circular motion at the point of contact is ɑ x r_{f} = Rɑ (k x j) = Rɑ i. For rolling without slipping, the acceleration of the CM must be opposite to the tangential acceleration, so A i = Rɑ i (third equation). We have three unknowns, f, A and ɑ, and three equations. After playing with algebra, I got: f = (T/3)(d  1/2) This means that if d = 1 (force on top), friction has the same direction as the tension; if d= 0 (force on center), friction is opposite to the tension; and if d = 1/2, there's no friction, so you could roll the cylinder on ice in this case. The fact that the direction of the frictional force changes based on how you apply the force is really weird, but I suppose it solves the problem, right? 


#5
Oct2910, 01:51 AM

HW Helper
P: 7,033

http://en.wikipedia.org/wiki/List_of_moments_of_inertia Still getting back to the original post, had that diagram shown a hollow cylinder, the friction force would be zero. It's not intuitive that the angular inertia and torque factors, translated into their linear equivalent, end up being relatively less than the linear inertia results in the friction force being in the same direction at the string (if the force is applied at r). 


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