Rotational motion question

Question

A uniform, solid cylinder with a mass M and radius R is pulled by a horizontal force F acting through the center. The cylinder rolls to the right without slipping. What is the magnitude of the force of friction between the cylinder and the ground?

Answer 1

The two central concepts one needs to understand in order to think about the situation described in the problem are 'Rolling without slipping' and the role of 'Static Friction'. Let us try to understand these one at a time.

Objects that have a round cross section (like a ball or a cylinder) can 'roll'. When an object is rolling, it experiences both translational motion of its center as well as rotational motion about its center. Correspondingly, it has both a velocity v (with which its center of mass is moving) as well as an angular velocity about its center, ω (measured in rad/s).

Since the object is undergoing two varieties of motion, so, in order to find the velocity of any point P on the object is the vector sum of the instantaneous velocities of P due to the two kinds of motion; v(P) = v_trans(P) + v_rot(P). The translational velocity v_trans(P) = v for every P on the body. However, the velocity of P due to rotational motion will depend upon the location of P. You can see by drawing a picture that v_rot(P) = ω r(P) in the direction ⊥ to the vector r(P) joining the center of the circle with P.

Now, a round object on a surface touches the surface at a single point. In an object which is 'rolling without slipping' this point is (momentarily) at rest (relative to the fixed reference frame of the lab). From the above, we can see that this will happen when v = ω R , where R is the radius of the object (because v_rot(P_touch) = ω R in direction opposite to v at the point P_touch where the object touches the surface). As a consequence, the displacements and accelerations are also related. If the object which is rolling without slipping moves by x, it has also rotated by θ = x / R. Similarly, when this rolling object accelerates (as will turn out to be the case in this problem), a = α R , where a is the linear acceleration, and α is the angular acceleration.

The other central thing to understand here is Static friction. Static friction is the form of friction experienced by a body which is at rest relative to the surface it is kept on, when there is a force attempting to slide it (but not yet succeeding). If you start pushing a block at rest, you cannot make it budge till you cross some minimum threshhold of force. What is keeping the block at rest is static friction. Like every other form of friction, static friction opposes motion. So, if I push a block with a force F (and it doesn't move), then the frictional force is f_s = F acting in the direction opposite to F. The maximum value that f_s can go up to is μ_s N, where N is the Normal (⊥) force experienced by the body from the surface, and μ_s is a number, called the 'coefficient of static friction'. In general, f_s in a given situation takes on whatever value is required to ensure that doesn't sliding doesn't happen, as long as that value is less than the maximum. If the required f_s to maintain the no slip condition is greater than the maximum possible, sliding starts.

Ok. Now, with the basics out of the way, let us think about the situation at hand. We have a cylinder which is being rolls to the right under the influence of force F. We wish to find the acceleration a. Let us list out the forces on it :

1) F, acting to the right at the center of mass

2) The gravitational force mg acting vertically down at the C.O.M.

3) The earth pulls the cylinder towards itself. As a result, the cylinder presses down on the surface. Newton's 3rd law : the surfaces presses up with a normal force N = mg pointing up, acting at P_touch

4) The frictional force f_s acting to the left at P_touch.

The forces in the horizontal direction lead to acceleration of the cylinder of mass M to the right with a, so :

Ma = F - f_s

Now, we need to analyze the rotational motion (that is, we need to write Newton's 2nd law for angular motion : τ_P = I_P α , where I_P is the moment of inertia of the object about the point P, and τ_P is the torque being applied about that point. The torque due to a force F applied at a point P a distance r(P) away the rotation axis we are considering is the product of r(P) with the component of F that is ⊥ to r(P). If you push on a wheel radially, it doesn't rotate. If you apply any force at the axle of the wheel, it doesn't rotate. In order to rotate it, you need to hold the rim and give it a tangential push. This is because the radial component of the force you are applying results in no torque about the axle, whereas the tangential component does.

About the center, the only force that exert a torque is the frictional force f_s. F and mg act at the center (thereby, r = 0, so τ = 0), whereas N acts in a line through the center (and so, has no tangential component. So,

τ = f_s R = I α

I , the moment of inertia takes the form I = k MR², where k is a factor ≤ 1 which shows how far from the rotation center the mass of the body is distributed. The smaller the value of k, the closer the concentration of the mass of the body to the axis of rotation under consideration. If all the mass is on the rim for a given round object (like in the case of a wheel or hollow cylinder), k = 1. For solid round objects, k < 1.

And finally, due to the no slip condition, α = a / R

So, putting everything together, we get two equations for two unknowns, f_s and a. These can then be solved to find a in terms of the knowns (M , R , k and F).

NOTE : Something to note about rotational motion is that we can stand on any point on a rotating object (so that it is still relative to us, and every other point on the object will be rotating relative to us). Let us say that when we are standing on the center of a wheel, the rim seems to rotate at angular velocity ω. Then, if we are standing at a point on the rim, the center will appear to rotate with the same velocity ω. So, rotation may be analyzed about any point. In this problem, two special points are the center and P_touch. Here, we have analyzed the angular motion about the center (because conceptually, the analysis is simplest about the center). You can try to perform this analysis about P_touch ; this will involve finding the moment of inertial I_touch about P_touch from the moment of inertia I_center about the center, using the 'parallel axis theorem', and figuring out which of the forces exerts a torque about P_touch. The results of this analysis should be cosnsitent with the solution of the above equations.

Hope this helps.