Rolling without slipping couples translational and rotational dynamics. The forces determine the center-of-mass acceleration, the torques determine the angular acceleration, and the no-slip constraint requires the two motions to remain compatible.

Consider an axisymmetric rigid body of mass $M$, radius $R$, and moment of inertia $I_{\rm CM}$ whose center of mass lies on its symmetry axis. It rolls down a straight incline of angle $\beta$ without slipping.

Along the slope, gravity contributes $Mg\sin\beta$. For passive rolling down the incline, static friction acts uphill and supplies the torque that increases the body's rotation. Newton's second law gives

$$Mg\sin\beta-f=Ma.$$

Taking torques about the center of mass, gravity and the normal force produce no torque about that point in this geometry, while friction gives

$$fR=I_{\rm CM}\alpha.$$

The no-slip condition gives

$$a=R\alpha$$

in magnitude. Combining the equations,

$$f=\frac{I_{\rm CM}}{R^2}a,$$

and therefore

$$a=\frac{g\sin\beta}{1+I_{\rm CM}/(MR^2)}.$$

This equation shows why mass distribution matters. A larger moment of inertia means that more of the motion is rotational, leaving a smaller center-of-mass acceleration for the same descent.

Shape matters, while mass and radius can cancel

Write

$$I_{\rm CM}=kMR^2.$$

Then

$$a=\frac{g\sin\beta}{1+k}.$$

For a uniform solid cylinder, $k=1/2$, so

$$a=\frac23g\sin\beta.$$

For a thin-walled hollow cylinder, $k=1$, so

$$a=\frac12g\sin\beta.$$

The solid cylinder therefore reaches the bottom first even if the two cylinders have the same mass and radius. In these idealized cases the acceleration depends on the dimensionless mass-distribution factor $k$, not separately on $M$ or $R$.

Static friction is whatever the constraint requires

From the torque equation,

$$f=kMa =\frac{k}{1+k}Mg\sin\beta.$$

Static friction is not automatically equal to $\mu_sN$. Rolling without slipping is possible only if the required friction does not exceed its limiting value:

$$f\le \mu_sN,$$

with

$$N=Mg\cos\beta.$$

Thus the surface must satisfy

$$\mu_s\ge\frac{k}{1+k}\tan\beta$$

for this incline problem. If that condition fails, the body slips and the no-slip equations no longer apply.

Friction does not always point opposite the center's motion

Friction opposes relative slipping or the tendency to slip at the contact point, not necessarily the motion of the center of mass. In this passive incline example it points uphill because, without friction, gravity would accelerate the center without producing the torque needed to satisfy the rolling constraint. Other applied forces or torques can reverse the required friction direction.

On a fixed surface under ideal pure rolling, the instantaneous contact point is at rest, so static friction does no mechanical work through that contact point even though it exerts a torque. The force can therefore be essential to the rotational dynamics without dissipating mechanical energy.