**Ball Rolling Down an Incline**

Let us place a sphere of uniform density with mass M and radius R on an inclined plane of angle θ. What is the distance travelled by the center of the ball as a function of time if it rolls down the plane without slipping? If the coefficient of static friction is μ, what is the maximum value of θ for which such rolling without slipping is possible?

The moment of inertia of a solid sphere is . We have the following forces acting on the sphere:

I. Gravity, F_{g}=Mg, downward, which can be treated as acting on the sphere’s center of gravity, which for the uniform gravity over the surface of the sphere is the center of mass, and thus the center of the sphere. In terms of components parallel and perpendicular to the incline, we have and , where the positive directions are down the incline for parallel and outward from the surface for perpendicular.

II. The normal force, which acts perpendicular to the plane, on the point of contact, and has magnitude such to make the net force perpendicular to the plane zero: .

III. Friction, F_{f}, which acts on the point of contact in the direction parallel to the plane and upward (negative parallel direction), with unknown magnitude f.

Newton’s second law (linear):

The net force, which acts in the direction parallel to the inclined plane, is . Via Newton’s second law,

Newton’s second law (angular):

We also need to consider angular motion of the sphere about the center of mass, as it is rolling. The only force producing a torque about the center of mass is the friction, which gives a torque of fR.

Thus the angular form of Newton’s second law gives

Solving these for the unknown friction and combining, we have:

.

Now, we consider what is needed for rolling without slipping. This requirement is that the angular velocity must ‘match’ the linear velocity at the rim: . For this to remain true, their derivatives must be equal as well, so .

Thus:

Note that this is a constant, so our distance as a function of time, assuming no initial velocity, is just

Our acceleration a here is 5/7 that of a mass M sliding down a frictionless incline of angle θ; the friction slows the linear motion. However, let us consider the energy at time t versus that the start. Letting the zero for the gravitational potential energy be at our starting height, we have . After time t, we have velocity , angular velocity , and gravitational potential energy . Thus we have energy

So the energy is conserved, and the friction serves only to transfer energy into the rotational motion.

Now, for rolling without slipping, the friction must be static friction, and thus the magnitude of the frictional force is less than the product of the coefficient of static friction times the magnitude of the normal force: .

Now, from our linear equation of motion,

.

Thus we have the inequality:

And thus, for us to have rolling without slipping, we require

Tags: Friday Physics, Science

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