5.6 Rolling Motion

One of the most familiar examples of torques in everyday life are rolling objects. Rolling motion is a superposition of a rotation about the centre of mass and a translation. In this section we will study rolling without slipping, which is the easiest type of rolling motion to analyse. We will also only look at circular rigid objects for now (wheels, balls, cylinders).

In this case, we have that the contact point with the ground has zero instantaneous velocity. For a circular rolling object, we can write

v_{\text{COM}}+v_{\text{rim}}=v_{\text{COM}}-R\omega=0.

(5.63)

This implies that

v_{\text{COM}}=R\omega,

(5.64)

which makes sense when we realise that for a circle that is rolling without slipping, its total distance travelled is $s=R\theta$ so $v_{\text{COM}}=\dot{s}=R\omega$ .

For a disc (or a cylinder), its moment of inertia while rolling is $\frac{1}{2}mR^{2}$ . Then its total kinetic energy is

$\displaystyle K$	$\displaystyle=\frac{1}{2}mv_{\text{COM}}^{2}+\frac{1}{2}I\omega^{2}$	(5.65)
	$\displaystyle=\frac{1}{2}mv_{\text{COM}}^{2}+\frac{1}{4}(mR^{2})\left(\frac{v_% {\text{COM}}}{R}\right)^{2}$	(5.66)
	$\displaystyle=\frac{1}{2}mv_{\text{COM}}^{2}+\frac{1}{4}mv_{\text{COM}}^{2}=% \frac{3}{4}mv_{\text{COM}}^{2}.$	(5.67)

So the rotational energy is half of its translational energy, or to put it another way a third of its kinetic energy is stored in the rotational motion.

5.6.1 Forces in Rolling Motion

Let’s look at the forces and torques involved in rolling. On a flat surface, if we have $a_{\text{COM}}=0$ then $\alpha=a/R=0$ . This implies that the torque about the centre of mass is zero which means no force is acting on the rim of the object. All this is to say that there is no tendency to slide. If $a_{\text{COM}}\neq 0$ , then there is a torque about the centre of mass and therefore there is a force. This must be a friction force, specifically static friction in the case of rolling without slipping. Thus we cannot have rolling without friction!

Which way does the static friction point? We are used to having static friction point to the left if the object is moving to the right, but there is a subtle gotcha here. In the case of rolling motion, the contact point wants to slip to the left, which means the static friction force must point to the right. On an incline, the weight does not exert a torque because it acts on the centre of mass, and the normal force does not exert a torque because it acts parallel to the radius vector. The only force which exerts a torque is the static friction, so $\tau_{f_{s}}=I\alpha$ .

Example 5.6.

A disc is rolling down an inclined plane. Find an expression for its acceleration.

Looking at the forces in the $x$ direction, we get

	$\displaystyle\sum F_{x}=ma_{\text{COM}}$	$\displaystyle=\lvert\boldsymbol{W}\rvert\sin\theta-\lvert\boldsymbol{f}_{s}\rvert$		(5.68)
		$\displaystyle=mg\sin\theta-f_{s}.$		(5.69)

Now we need to find an expression for the magnitude of static friction $f_{s}$ , which we can do by analysing the torques. Since the only force which exerts a torque is static friction, we have $\tau_{f_{s}}=Rf_{s}=I\alpha$ . So using the moment of inertia for a disc $\frac{1}{2}mR^{2}$ and substituting $\alpha=a_{\text{COM}}/R$ , we get

f_{s}=\frac{Ia_{\text{COM}}}{R^{2}}=\frac{1}{2}ma_{\text{COM}}.

(5.70)

Substituting this into the equation above, we get

	$\displaystyle ma_{\text{COM}}$	$\displaystyle=mg\sin\theta-\frac{1}{2}ma_{\text{COM}}$		(5.71)
		$\displaystyle\implies a_{\text{COM}}=\frac{2}{3}g\sin\theta.$		(5.72)

This result is somewhat interesting, the acceleration does not depend on the mass or radius of the disc, only the angle of the plane. However, the acceleration does depend on the distribution of mass within the rolling object, i.e. the moment of inertia. If we had left the moment of inertia unspecified, we would get

a_{\text{COM}}=\frac{mg\sin\theta}{\frac{I}{R^{2}}+m}.

(5.73)

This is valid for all circular rolling objects. As we have seen, the moment of inertia for most circular objects is proportional to $mR^{2}$ , leading to a simplification with a different constant multiplying $g\sin\theta$ .