Physics Puzzle: is Rolling Possible?
In the middle of a discussion on rigid bodies in a Mechanics class given by a seasoned physics professor, a freshman physics student suddenly stands up and says, “it is impossible for a sphere to roll on a plane.” When asked to explain his statement, he argues as follows:
Consider a sphere rolling on a plane, with angular velocity ω about the center-of-mass and a center-of-mass velocity V. For rolling without slipping, one must have V=ωR, where R is the sphere’s radius. There are three forces acting upon the sphere: the normal force, gravity and friction at its contact point with the table. Gravity and the normal force “cancel out”, in the sense that they do not affect the ball’s horizontal motion, nor do they apply any torque on it, and thus the only force that has to do with rolling is friction. According to Newton’s laws, and their rigid-body generalizations, one has:
M * (dV/dt) = -f
for the center of mass’s motion, and:
I * (dω/dt) = -f * R
for its rotational motion about the center of mass, where R is the sphere’s radius, M its mass and and I its moment of inertia:
Above: a sphere rolling on a table. Below: friction's action on the sphere.
For rolling without slipping, dV/dt = R * dω/dt. Substituting this into the first equation and multiplying both sides by R yields:
MR²*(dω/dt) = -fR
With the right hand side of both equations equal, we can equate the left-hand-sides:
MR²*(dω/dt) = I * (dω/dt)
This seems to imply that I=MR². But everybody knows that a sphere’s moment of inertia is really (2/5)*MR² - a contradiction! Hence, rolling without slipping is an impossibility.
How did the physics professor reply to the freshman’s claims?
The normal force only cancels the component of gravity perpendicular to the surface.
“fuck you”
If I had to guess, it has to do with the fact that this formulation does not take into account the actual distribution of mass withing the sphere. In fact, these equations hold perfectly true for other shapes like a thin hoop or circle rotating about its radial axis. The equations are a generalization for any rolling object. Odds are, the physics professor or someone more skilled with physics than I am would know where and why to add in a constant factor that takes into account the distribution of mass in the object.
Since there is no slipping, the angular velocity is constant. Therefore dw/dt is 0, and to reach the final line, you need to divide by 0… OH SHI-
I believe the professor would respond with a quick slap to the face of this student for completely misunderstanding the concept of friction (as well as interrupting his discussion of rigid bodies). The student has friction pointed in the wrong direction! This sad excuse for a free body diagram would leave the professor appalled. At the point where the sphere touches the plane, the velocity vector points to the LEFT and therefore the friction force must be pointing to the RIGHT. Don’t get me started on the fact that he’s working with three dimensional shapes and two dimensional equations. The sum of the moments does not equal I(dw/dt) it equals (dH/dt) in three dimensions. Sit down student, and start paying attention in class instead of doodling.
Please, let’s leave the verbal and physical abuse out of it, a physics student’s life is hard as it is!
A Gamer got it right. It’s a fallacious proof which uses division by 0. A good lesson in watching out for hidden assumptions!
Aaron, gravity is perpendicular in its entirety to the surface. It has no parallel component (well, at least not in this problem. But even in real life, these non-perpendicular components are extremely small and wouldn’t go as far as to affect a ball’s motion considerably).
Anonymous, the given moment of inertia does take into account the mass “inside” the sphere (assuming it’s homogeneous).
Dane, the problem can be formulated completely in two dimensions because of its symmetry. I’m not sure what H is in your equation, but if it’s angular momentum, then - unless the moment of inertia changes with time - it’s equal to I*dw/dt identically.
nicely put, Assaf.
non-slipping friction on a rolling object involves static friction at the point of contact since its instantaneous velocity is zero. i found the following to be a nice little review of the stuff: http://dept.physics.upenn.edu/courses/gladney/mathphys/java/sect4/subsubsection4_1_4_3.html
Icepaxx was answering the question. He thinks the Professor’s response was “fuck you”.
I think the Professor circled the X and said “There it is!” before drawing an elephant and declaring that the elephant pushed the sphere to make it roll. But I could be wrong, I wasn’t there you see.
My first thought was that the sphere had to be set in motion somehow. Assuming that whatever set the ball in motion left the area of observation with equal angular momentum, the ball wouldn’t travel anywhere.
I’m with Gamer on this one. Dane, if the friction force was pointing to the right, it would be exerting a torque opposite to the rotation. This would mean the ball is either slowing down, or is rotating counter-clockwise while moving right.
There are two ways to solve the problem.
One way is to write the correct equation of motion
M dv/dt+ I d omega/dt=F and solve it with the assumption omega*r=v.
It can be easily seen from an energy argument that that the change in rotation + translation kinetic energy is equal to the work done by friction.
There are two ways to solve the problem.
One way is to write the correct equation of motion
M dv/dt+ Ir d omega/dt=F and solve it with the assumption omega*r=v.
It can be easily seen from an energy argument that that the change in rotation + translation kinetic energy is equal to the work done by friction.
Well, first of all, dv/dt =omega*omega*R.
Secondly, rolling without friction does not imply any force! So, if F=0, dv/dt is also zero, so d(omega)/dt is zero too. This implies v=const and omega=const. The ball rolls without sliding if v=omega*R, else, sliding occurs and a force of friction appears.
This proof is completely incorrect. Since the sphere rolls without slipping, there cannot be any external forces acting on it. In the above statement, there is a force of friction exerting on the sphere. It means that, when we assume the spere as if it were a point particle, it constantly slows down because of force of friction. And after some time, it stops. Because of the fact that rolling without slipping is continuous motion, there cannot be such a situation mentioned above. Furthermore, another aspect is that, when we consider the contact point, we can see that there is acceleration toward left and that’s all! We need at least a force which can cause an acceleration toward right such that the sphere is able to roll without slipping. As you probably know, in rolling without slipping, the acceleration of the contact point has to be zero.
“Since the sphere rolls without slipping, there cannot be any external forces acting on it.” It is a wrong sentence. If an object rolls without slipping in a horizontal plane and unless you apply a force on it, there is no force of friction. The basic idea is that the acceleration of the contact point has to be zero in rolling without slipping with respect to an outside observer. Therefore, “It is impossible for a sphere to roll on a plane.” is invalid!