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# Laws of rolling motion

Module by: Sunil Kumar Singh. E-mail the author

Summary: Pure rolling is governed by Newton's laws as applicable to pure translation and pure rotation.

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A general reference of Newton's laws of motion dictates that a disk rolling along a straight path should continue rolling with same velocity unless acted upon by the external force. If there is a net external force acting on the body, then it will either speed up or slow down, depending upon the direction of force and torque with respect to the direction of motion. We shall discuss these aspects pertaining to the dynamics of the rolling motion in this module. However, we shall restrict mathematical formulations of laws governing rolling motion to pure rolling (also referred as smooth rolling) only. This means that external forces, being considered here for formulating laws of rolling motion, should not induce sliding of the rolling disk.

## Laws governing pure rolling motion

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 :

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 :

v C t = ( ω t ) R a C = α R v C t = ( ω 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.

## Rolling motion and friction

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.

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. The golden rule to avoid this confusion is simple. Identify the direction of net force (excluding friction). Friction shall work in the opposite direction to that of the net force. Because, it tends to counteract the rolling disk to slide. The rolling disk, on the other hand, moves in the direction of net force. This is the case when a disk or cylinder rolls down an inclined ramp. Here, component of gravitational force is the net force moving the disk down the ramp. The friction, therefore, is in the opposite direction to that of component of gravitational force.

There are, however, situations in which a system moves from internal mechanism as available with bicycle or a car. As a matter of fact, the mechanism allows to convert internal force into external force in the form of friction force. Let us consider the case of a bicycle wheel. When we paddle to accelerate the bicycle, the rolling friction comes into picture. This friction is actually responsible to accelerate the bicycle, because this is the only external force ! Evidently, friction here works in the direction of the motion of the rolling wheel. In case, we apply break to decelerate the bicycle, the friction works in the opposite direction to that of motion.

In all situations, the direction of friction is such that it opposes the tendency of sliding that results from application of force.

## Rolling without sliding along an inclined ramp

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 :

1. Force of gravity, mg, acting downward.
2. Normal force, N, perpendicular to the ramp in y-direction.
3. Static friction, f s f s , acting upward.

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.

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 :

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

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