Corresponding to two motion types involved with pure rolling, there are two laws governing the motion. One governs the linear motion of the center of mass, whereas the other governs the rotational motion of the rolling disk.

For pure translation of the center of mass, the second law of Newton's law is :

F = m a C F = m a C

Here, "F" denotes the resultant force in the direction of motion and " a C a C " denotes the linear acceleration of the center of mass of the disk. Further, we should note that we are discussing motion in one direction and shall, therefore, use the scalar form of the law of motion as given above.

For pure rotation of the disk about an axis passing through the center of mass and perpendicular to its surface, the second law of Newton's law is :

t = I α t = I α

where "I" denotes the moment of inertia and "α" denotes angular acceleration about the axis of rotation through the center of mass.

If the rolling is pure i.e. without sliding, then angular acceleration can be related to linear acceleration of the center of mass. In the previous module, we developed this relation between linear and angular velocities as :

Figure 1: The angular and linear velocities are related to each other in pure rolling.

Pure rolling motion

v C = ω R v C = ω R

where "R" is the radius of the rotating disk. It must be kept in mind that this relation was developed for the case of pure rolling. As such any derivation based on this relation will be valid only for pure rolling that does not involve sliding. Now, differentiating this equation with respect to time, we have :

ⅆ v C ⅆ t = ( ⅆ ω ⅆ t ) R ⇒ a C = α R ⅆ v C ⅆ t = ( ⅆ ω ⅆ t ) R ⇒ a C = α R

Three equations namely (i) Newton's law of motion for pure translation (ii) Newton's law of motion for pure rotation and (iii) relation connecting linear and angular acceleration, determine the dynamics of pure rolling motion.

Friction has presence everywhere in our day to day life. It is important to understand how friction affects a rolling motion. A pure rolling disk with constant velocity does not slide on the surface. In this case, there is no tendency of the rolling disk to slide on the surface in contact. Ideally, there is only a point contact with the surface underneath. As such, there is no friction between surfaces.

Absence of friction for rolling at constant velocity has a very significant implication as a disk in pure rolling shall move indefinitely if no other force is acting. This is a slightly unrealistic deduction for we know that all rolling disk is brought to rest ultimately unless external force is applied to maintain the speed. This needs explanation.

As a matter of fact, it is not possible to realize an ideal pure rolling in the first place. All rolling motion in our real world involves contact which spreads beyond a point and there is some amount of sliding involved and therefore some amount of friction is also present to decelerate the rolling. Nevertheless, friction of rolling is comparatively much smaller. This is the reason that all transportation modes (car, rail etc.) on the surface are almost always designed to have rolling wheels.

Figure 2: All rolling motion in our real world involves contact which spreads beyond a point.

Pure rolling motion

When external force is applied it changes the linear and angular velocities in accordance with the relative direction of external force (in the direction of motion or opposite to it). The disk begins to move either faster or slower. If the disk still rolls without sliding, then it is not possible that there is an increase in one component, say linear velocity, and a decrease in other component i.e angular velocity. It means that the external force can not selectively change linear or angular component in pure rolling. It is so because two velocities are bound by the relation :

v C = ω R v C = ω R

So if linear velocity increases, then the angular velocity also increases and vice-versa.

The change in velocities, however, induces a "tendency" to slide. The point in contact tends to slide as against rolling. This is the first implication of an external force on the rolling disk. If the disk rolls without sliding even after application of external force, then the friction on the rolling disk is static friction. The static friction can have values in the range from zero to maximum static friction ( μ s N μ s N ).

f s < μ s N f s < μ s N

However, if the rolling disk begins to slide as a result of the application of net external force (this happens when we give sudden acceleration or deceleration to the rolling disk), then friction at the contact is kinetic friction ( μ k N μ k N ).

f k = μ k N f k = μ k N

Under this situation, the rolling is no more pure as the rolling disk slides on the surface. The correspondence between angular and linear velocities as valid for pure rolling is not valid now. This means that :

v C ≠ ω R v C ≠ ω R

We can appreciate this domain of rolling motion in the context of our day to day experience. Consider applying brakes on a fairly smooth wet road, while driving a car. The breaking pad housed on the axle decreases angular velocity (ω), while the car skids on the road over a distance before coming to rest. If the brake applied is sudden and hard, then the wheel might hardly rotate, whereas it slides with the velocity that of center of mass along the road. Here,

v C > ω R v C > ω R

In this case, center of mass moves greater than "2πR" by the time the wheel completes one rotation. There may be an opposite situation. We may think of accelerating fast on a slippery wet road. The wheel rotates faster, whereas the car as a whole moves not so fast. Here,

v C < ω R v C < ω R

In this case, center of mass moves lesser than "2πR" by the time the wheel completes one rotation.

Direction of friction

Direction of friction is sometimes a source of confusion. To analyze the situation, let us examine the acceleration of rolling disk due to external agency. We can accelerate rolling by applying force through its center of mass. Such forces do not constitute a torque about axis of rotation and has the tendency to impart sliding in the direction of acceleration of COM. However, static friction force comes into picture counteracting sliding tendency due to external force so that disk rolls without sliding. Evidently, friction force has to act in opposite direction to the component of external force parallel to contact surface. As angular acceleration is tied to linear acceleration in rolling, the disk begins to rotate faster i.e disk is accelerated rotationally as well. Rolling of disk over an incline is such case. The external force acts through center of mass. As such, friction acts in the opposite direction to that of net force through center of mass parallel to the contact surface. And the disk rolls faster. The friction here works in the direction opposite to that of the acceleration of the rolling wheel. Also note that friction, in this case, tends to impart a torque that increases angular acceleration.

Figure 3: Friction acts in the direction opposite to that of net external force along the contact.

Direction of friction

Another way to accelerate rolling is to impart a force such that its line of action does not pass through center of mass. In this case, force constitutes a torque that impart angular acceleration. This makes the disk to slide in the opposite direction to the direction of acceleration of COM. Again, friction force comes into picture counteracting this tendency. Evidently, friction force has to act in the direction of the component of external force parallel to contact surface. Alternatively, the friction force acts constituting an opposite torque to the one which imparts angular acceleration to it. Since angular acceleration is tied to linear acceleration of the center of mass, the disk translates faster as well. The rolling of a bicycle wheel is such case. When we paddle to accelerate the bicycle, we impart angular acceleration. The friction here works in the direction of the acceleration of the rolling wheel. Also note that friction in this case tends to impart a torque that decreases angular acceleration. In case, we apply break to decelerate the bicycle, the friction again works in the direction of acceleration (deceleration or negative acceleration).

Figure 4

Direction of friction
(a) Acceleration of motion : Direction of friction is same as that of acceleration. (b) Deceleration of motion : Direction of friction is same as that of acceleration.

In all situations, the direction of friction is such that it opposes the tendency of sliding that results from application of force. If the rolling is accelerated by net force through COM, then friction is opposite to the direction of acceleration. The torque due to friction increases angular acceleration. On the other hand, if the rolling is accelerated by a force constituting torque, then friction is in the direction of acceleration. The torque due to friction decreases angular acceleration.

We consider rolling down of a rolling disk along an incline. We select an appropriate pair of rectangular coordinate system such that motion is along the positive direction of the x-coordinate. The various forces acting on the rolling disk are :

Figure 5: Forces acting on the rolling disk.

Rolling down the incline

Confining force analysis in x-direction, we apply Newton's law of motion for linear motion as :

∑ F x = mg sin θ - f s = m a C ∑ F x = mg sin θ - f s = m a C

Similarly, we apply Newton's law for rotation :

t = I α t = I α

We note here that force due to gravity and normal force pass through the center of mass. As such, they do not constitute torque on the rolling disk. It is only the friction that applies torque on the disk, which is given as :

t = f s R = I α t = f s R = I α

At this stage, we can make use of third equation that connects linear and angular accelerations for rolling without sliding :

a C = α R a C = α R

Combining this equation in the expression of torque,

f s R = I α = I a C R ⇒ f s = I a C R 2 f s R = I α = I a C R ⇒ f s = I a C R 2

Substituting the expression of friction in the equation of force in x-direction, we have :

mg sin θ - f s = m a C ⇒ mg sin θ - I a C R 2 = m a C ⇒ a C ( m + I R 2 ) = mg sin θ ⇒ a C = g sin θ ( 1 + I m R 2 ) mg sin θ - f s = m a C ⇒ mg sin θ - I a C R 2 = m a C ⇒ a C ( m + I R 2 ) = mg sin θ ⇒ a C = g sin θ ( 1 + I m R 2 )

We can use this expression to find linear acceleration (hence angular acceleration also) for circular rolling bodies like ring, disk, cylinder and sphere etc. We only need to use appropriate expression of moment of inertia as the case may be.

Example 1

Problem : A long rope of negligible mass is wrapped many times over a solid cylinder of mass "m" and radius "R". Other end of the rope is attached to a fixed ceiling and the cylinder is then let go at a given instant. Find the tension and acceleration of the cylinder.

Figure 6: The rolling of the cylinder is guided on the rope.

Rolling down the rope

Solution : The rolling of the cylinder is guided on the rope. This is similar to rolling on an incline. We select a coordinate system in which y-direction is along the downward motion. Following the same steps of analysis as described earlier :

Figure 7: Free body diagram.

Rolling down the rope

From the application of Newton's second law for translation in y - direction :

∑ F y = mg - T = m a C ∑ F y = mg - T = m a C

From the application of Newton's second law for rotation :

t = T R = I α t = T R = I α

Note that force due to gravity passes through center of mass and does not constitute a couple to cause angular acceleration. Now, from relation of linear and angular accelerations, we have :

a C = α R a C = α R

Putting in the equation of torque, we have :

T = I a C R 2 T = I a C R 2

The moment of inertia for solid cylinder about its axis is,

I = m R 2 2 I = m R 2 2

⇒ T = m R 2 a C 2 R 2 ⇒ T = m a C 2 ⇒ T = m R 2 a C 2 R 2 ⇒ T = m a C 2

Substituting the expression of friction in the equation of force analysis in y-direction, we have :

mg - m a C 2 = m a C ⇒ g - a C 2 = a C ⇒ a C ( 1 + 1 2 ) = g ⇒ a C = 2 g 3 mg - m a C 2 = m a C ⇒ g - a C 2 = a C ⇒ a C ( 1 + 1 2 ) = g ⇒ a C = 2 g 3

Putting this value in the expression of "T", we have :

⇒ T = m a C 2 = m g 3 ⇒ T = m a C 2 = m g 3