 # Rolling motion, formula with examples to know.

During our day-to-day activities, we often encounter the rolling motion. Rolling motion is a type of motion that includes the combined motion of rotational and translational motion of an object with respect to a surface in ideal conditions, where no sliding happens

## Rolling motion,

During our day-to-day activities, we often encounter the rolling motion. Rolling motion is a type of motion that includes the combined motion of rotational and translational motion of an object with respect to a surface in ideal conditions, where no sliding happens.

An object like a disc, wheel, ball or any circular body that is rolling over a stationary ground surface undergoes rolling motion and it is an important point to note that at every instant, there is a single point of contact without slipping.

Why do such things happen? Because at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. Where the point of contact with the ground has zero velocity, which matches the ground velocity and is not slipping.

For an object, the motion of the center of mass is the motion of that object. The center of mass is a position defined relative to an object or system of objects. During the rolling motion of an object, the surface in contact gets deformed a little bit temporarily.

Due to this deformation, a finite area of both the object and surface comes together. Then what is the effect of this phenomenon on the object and the surface? The effect is that the components of the contact force parallel to the surface oppose motion resulting in friction.

Let us discuss how the rolling motion works. Suppose vcm is the velocity of the center of mass and is the translational velocity of the disc-like object. As the center of the rolling disc is at its geometric center C, vcm is the velocity of C (Image A), which is parallel to the level/rolling surface.

The rotational motion of the object happens about its axis of symmetry, which passes through C. Thus the velocity of any point of the object like p0, p1, or p2, consists of two parts, one is the translational velocity  vcm and the other is the linear velocity vr and due to rotation, the magnitude of vr is rω i.e vr=rω, where

• ω is the angular velocity of the rotation of the disc-like object about the axis.
•  r is the distance of the point from the axis (i.e from C)

The velocity vr is perpendicular to the radius vector of the given point with respect to C.  At point P0 of image A, vr is exactly opposite to vcm due to rotation, where the magnitude of

vr is Rω. where

• R is the radius of the disc-like object.

Hence the condition for the disc-like object to roll without slipping is

vcm=Rω.

At P1 there are two velocities exist, i.e vcm and vr. Now at P1 the top of the object, the magnitude of velocity  is,

v1=vcm+vr=vcm+Rω=Rω+Rω=2Rω or vcm+vcm=2vcm or

v1=2Rω=2vcm

Such a condition is applicable to all rolling bodies.

## The kinetic energy of the body in a rolling motion

The kinetic energy of a rolling body can be separated into the kinetic energy of translational motion and the kinetic energy of rotational motion. The kinetic energy of the translational motion is given by the following formula.

#### Formula

KT=mv2cm/2, where

• m is the mass of the rolling body.
• vcm is the center of the mass velocity.

Further, the motion of the rolling body about the center of mass in rotation, KR represents the kinetic energy of the rotation of the body;

KR=Iω2/2, where

• I represent the moment of inertia about the axis of symmetry of the rolling body.
• ω is the angular velocity.

Now the kinetic energy of the pure rolling body is given by the formula;

#### The formula of the kinetic energy of the pure rolling body;

K=1/2Iω2+1/2 mv2cm where

• m is the mass of the rolling body.
• vcm is rotation motion.
• I represent the moment of inertia.
• ω is the angular velocity of the rolling body.

Then the kinetic energy in terms of the radius of the gyration of the body is given by the equation;

K=1/2 mv2cm(1+k2/R2), where

• k is the corresponding radius of gyration.

## Rolling motion on an inclined plane

Let us see image B, a rigid body with radius R is rolling down an inclined plane at an angle θ with the surface. When the said body is placed on an inclined plane tries to slip down, which leads to static friction force that acts upward. The torque is produced due to friction, which rotates the body. Image of a rolling motion on an inclined plane./ Image B

Now consider the linear acceleration of the center of mass as a and angular acceleration as α and the radius of gyration is k. From the above, figures the linear motion of the object  is mg sin θ-f=ma—–1

For rotational motion net torque is fR is given as Iα, where I represent the moment of inertia about the axis. The equation is;

fR= Iα

fR=(mk2) α———2

As there is no slipping, it is pure rolling and at the point of contact, the body is at rest. Now the condition for pure rolling is v=Rω.

Then the equation for linear acceleration a=Rα  ———  3

After solving the above three equations,1,2 and 3 for α and f, we get the following formula,

#### Formula;

α =gsin θ/1+k2/R2

f =mg sin θ/1+ k2/R2

We may also derive the condition for pure rolling, to avoid slipping,

f  ≤  µsN

g sin θ/1+k2/R2   ≤  µs mg sin θ

µs ≥ tan θ/1+k2/R2

Due to the above condition for µs, the body rolls without slipping.

## Examples of rolling motion;

The examples of rolling motion in our day-to-day life are observed frequently. Some of the examples are;