Question

What conditions must be met for a ball to roll perfectly down an incline without slipping?

Answer 1

The torque provided by the friction must be sufficient to enable to ball to have sufficient angular acceleration so that its rotation keeps up with the movement of the center of mass.

It's an interesting problem. Work it through (which I haven't, yet) and you'll find some sort of condition on the coefficient of static friction as a function of many of the other variables (coeff of static friction, incline of plane, radius of ball, ...; I wonder if the mass drops out, I hypothesize it does.)

Answer 2

I think we're assuming that the ball begins with some angular velocity great enough to escape the high bound of static friction. But, uh, I really have no idea how to set this up.

Answer 3

Either that or the incline just has to be sufficient, even if it's starting at rest. I'll have a look at this problem tomorrow morning and try and write it out.

Answer 4

So first, a diagram:

Answer 5

Now the strategy in solving this problem is this: there are two ways to calculate the acceleration of the body. One is using translational, basic kinematics: how fast does the ball roll down the incline, ignoring the possibility of it spinning? Call that acceleration the linear acceleration \( a_L \). And then how fast is the ball rotating given the torque applied to the ball, call that \( a_R \), R for rotational. Now if the ball doesn't skid, then we want
\[ a_R \geq a_L \] and this will give us a constraint on the coefficient of friction \( \mu \).

Answer 6

Let's calculate \( a_L \) first as this is straight forward. There are two forces acting in the direction of the plane: the component of gravity and the frictional force, hence the net force in that direction is
\[ mg \sin \theta - mg \mu \cos \theta \] and thus
\[ a_L = g \sin \theta - \mu \cos \theta .\]

Answer 7

Calculating \( a_R \) is less obvious. If the ball is not slipping, then at the point of contact between the ball and the plane, it is rotating. Let's calculate torque around the point of contact.

At that point, the force of friction, \( F_f \) has no torque. The normal force \( N \) also has no torque because it is in the direction of the displacement vector at that point, the center of mass. Hence the only force at play is the gravitational force and the torque from that force is
\[ \tau = rmg \sin \theta \]
Now torque \( \tau = I \alpha \). For a sphere, the moment of inertia through an axis through its center is \( I_c = \frac{2}{5}mr^2 \). Using the parallel axis theorem, the moment of inertia of the ball at the point of contact is
\[ I = I_c + mr^2 = \frac{7}{5}mr^2 \]
Hence
\[ \tau = I \alpha \implies rmg \sin \theta = \frac{7}{5}mr^2 \alpha \] and therefore
\[ \alpha = \frac{5g \sin \theta}{7r} \]
This is the angular acceleration about the point in contact. At the center of mass therefore, the linear acceleration is
\[ a_R = r\alpha = \frac{5g \sin \theta}{7} \]

Answer 8

Now we are ready to compare the two. If the ball does not slip, then
\[ a_R \geq a_L \]
That is, canceling \( g \) from both sides
\[ \frac{5}{7} \sin \theta \geq \sin \theta - \mu \cos \theta \]
Rearranging we find
\[ \mu \geq \frac{2}{7} \tan \theta \]

This is quite pleasing, as the final answer is independent of \( g \), \( r \) and mass \( m \). Further, as \( \theta \) increases, so does \( \mu \) which makes intuitive sense and indeed as \( \theta \rightarrow \pi/2 \), the perpendicular, if the ball doesn't slip, then \( \mu \rightarrow \infty \).

Answer 9

the angle between radius vector and mgsin@(acting radially is 0)
so we can see that there is torque to cause rolling motion only by frictional force
there shud be an essential torque to initial rolling motion according to newtons 1rst law of motion
this is my part of the explanation which just adds a tiny speck to james's ellaboration up here.