System of Particles and Rotational Dynamics

Rolling Motion

Every one of you must have seen a ball rolling down a hill or rolling of bike wheels when the bike is moving. You must have also seen the motion of a bowling ball or a motion of snooker ball on the table. These are nothing but the examples of a rolling motion.  Now let us study this in detail.

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Rolling Motion

Let us understand the concept of  “Rolling Motion”. Imagine a box sliding down from an inclined plane. This is an example of a translation motion. They have the same velocity at an instant of time. The motion of a ceiling fan or a merry go round are the examples of rotational motion. Here in rotational motion, every particle of the body moves in a circle. Rolling motion is the combination of rotation and translation.

For example, an object, say a ball is in rolling, that is it is rolling on the surface of the ground. So the ball is in the rotation motion. At the same time, the ball is moving from one point to another point so there is a translation motion. At any instant of time, there is one point which is always in contact with the surface. So that point is at rest.

Rolling Motion of Disc

                                                                                                                                            (Source: My Rank)

Let us assume that the disc rolls without slipping. If that this disc is the uniform disc the centre of mass lies at the centre of the disc that is the point ‘0’.  The velocity of the centre of mass will always be parallel to the surface as we can see this in the figure.

At any instant of time, there will be two velocities. One is the velocity of the centre of mass and another one is the component of linear velocity. The velocity of the centre of mass is vcm. This corresponds to the translation motion of the object. vis the linear velocity which corresponds to the rotation.
v= rω.

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Let us first consider the point P0

So at Pwe have two velocities that are vcm and vr. Here the direction of the centre of mass is in the direction of Pand vcm and vare opposite in direction. If we want the object to roll without slipping, the point which is in contact with the ground should be at rest. So vr = vcm 

⇒ rω = vcm

The above is the condition for rolling without slipping.

At point P1

Also here there are two velocities that are vcm and v1. So the net velocity of P is,

v=  vcm +  v=  vcm + rω = rω + rω = 2rω

The same is true for all the points on the disc.

Kinetic Energy for Rolling Motion

Since rolling is the combination of rotational motion and translational motion,

K.E = K+ KR

= \( \frac{1}{2} \) mv² + \( \frac{1}{2} \) Iω²

= \( \frac{1}{2} \) m vcm² + \( \frac{1}{2} \) Iω²

I = moment of inertia. We can write moment of inertia as I = mk², k = radius of gyration.

= \( \frac{1}{2} \) m vcm² + \( \frac{1}{2} \) mk² ω²

For rolling without slipping the mathematical condition is rω = vcm

K = \( \frac{1}{2} \) m² ω² + \( \frac{1}{2} \) mk² ω²

= \( \frac{1}{2} \) m vcm² + \( \frac{1}{2} \) mk² \( \frac{v_cm²}{r²} \)

⇒ K = \( \frac{1}{2} \) m vcm² [ 1 +  \( \frac{k²}{r²} \) ]

This is the kinetic energy of a rolling  motion.

Questions For You

Q. Sand Sare two spheres of equal masses.  Sruns down a smooth inclined plane of length 5 cm and of height 4 cm and S2 falls vertically down by 4 cm. the work was done by all forces acting on S1

  1. Is greater than that on  S
  2. and on S are same and non zero
  3. is less than that of  S
  4. as well as on  S2  is zero.

Solution: B. Using energy balance we see that potential energy which only depends on the height changes to kinetic energy at the bottom. Thus the work done in both the cases is same. If friction were present work done would have been different.

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