What is the difference between rolling and slipping?

Rolling motion involves both translation and rotation. Pure rolling, specifically, occurs when there is no relative motion between the point of contact of the rolling body and the surface it's rolling on. This means the instantaneous velocity of the contact point is zero. Slipping, on the other hand, occurs when there *is* relative motion at the contact point. If the body slides forward while rotating, it's forward slipping. If it rotates but doesn't move forward enough, it's backward slipping or skidding. The key distinction is the 'no-slip' condition ($v_{CM} = Romega$) for pure rolling.

Does friction do work in pure rolling motion?

No, static friction does no work in pure rolling motion. While static friction is crucial for providing the torque necessary for the body to rotate and thus roll, the point of application of this static friction force is instantaneously at rest relative to the surface. Since work done is force times displacement in the direction of force, and the displacement of the point of application is zero, the work done by static friction is zero. It merely converts potential energy into kinetic energy (both translational and rotational).

Why does a solid sphere roll faster than a hollow sphere down an inclined plane?

The acceleration of a body rolling down an inclined plane is given by $a_{CM} = \frac{g\sin\theta}{1 + \frac{I_{CM}}{MR^2}}$. A solid sphere has a moment of inertia $I_{CM} = \frac{2}{5}MR^2$, so $I_{CM}/MR^2 = 2/5 = 0.4$. A hollow sphere has $I_{CM} = \frac{2}{3}MR^2$, so $I_{CM}/MR^2 = 2/3 \approx 0.67$. Since the hollow sphere has a larger $I_{CM}/MR^2$ value, its denominator $(1 + I_{CM}/MR^2)$ is larger, resulting in a smaller acceleration. This means the solid sphere, with more mass concentrated closer to its axis of rotation, accelerates faster.

What is the instantaneous axis of rotation in pure rolling?

In pure rolling motion, the point of the rolling body that is instantaneously in contact with the surface is at rest relative to the surface. This point acts as the instantaneous axis of rotation (IAOR). All other points on the rolling body can be considered to be rotating about this IAOR at that specific instant. This concept simplifies velocity calculations, as the velocity of any point can be found by $v = r_{IAOR}omega$, where $r_{IAOR}$ is its perpendicular distance from the IAOR and $omega$ is the angular velocity of the body.

How is the total kinetic energy of a rolling body calculated?

The total kinetic energy of a rolling body is the sum of its translational kinetic energy and its rotational kinetic energy about its center of mass. Mathematically, it is expressed as $K = \frac{1}{2}Mv_{CM}^2 + \frac{1}{2}I_{CM}\omega^2$. Here, $M$ is the mass, $v_{CM}$ is the velocity of the center of mass, $I_{CM}$ is the moment of inertia about the center of mass, and $omega$ is the angular velocity. For pure rolling, we can use the condition $v_{CM} = Romega$ to express the total kinetic energy entirely in terms of $v_{CM}$ or $omega$, for example, $K = \frac{1}{2}v_{CM}^2 (M + \frac{I_{CM}}{R^2})$.

Rolling Motion

Physics

NEET UG

Version 1Updated 22 Mar 2026

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Rolling motion is a complex form of rigid body motion that combines both translational and rotational motion simultaneously. For an object to undergo pure rolling, there must be no relative motion between the point of contact on the rolling body and the surface it is rolling upon. This 'no-slip' condition is crucial and implies a specific relationship between the linear velocity of the center of m…

Quick Summary

Rolling motion is a fundamental concept in physics, representing a combination of translational and rotational motion. For 'pure rolling' (without slipping), the crucial condition is that the point of contact between the rolling body and the surface is instantaneously at rest.

This leads to the kinematic relationship $v_{CM} = Romega$ , where $v_{CM}$ is the velocity of the center of mass, $R$ is the radius, and $omega$ is the angular velocity. The total kinetic energy of a rolling body is the sum of its translational kinetic energy ( $\frac{1}{2}Mv_{CM}^2$ ) and rotational kinetic energy ( $\frac{1}{2}I_{CM}\omega^2$ ).

When a body rolls down an inclined plane, static friction provides the necessary torque for rotation but does no work. The acceleration of the center of mass depends on the body's moment of inertia, with objects having smaller $I_{CM}/MR^2$ accelerating faster.

Understanding the moment of inertia for various shapes is key to solving problems related to rolling motion.

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Key Concepts

Pure Rolling Condition (

v_{CM} = Romega

)

This condition is the cornerstone of pure rolling. It means that the linear speed at which the center of the…

Total Kinetic Energy of a Rolling Body

A rolling body possesses both linear and rotational motion, and thus its total kinetic energy is the sum of…

Acceleration on an Inclined Plane

When a body rolls down an inclined plane without slipping, its acceleration is less than $g\sin\theta$ (which…

Pure Rolling Condition: —

v_{CM} = R\omega

Total Kinetic Energy: —

K = \frac{1}{2}Mv_{CM}^2 + \frac{1}{2}I_{CM}\omega^2

Acceleration on Incline: —

a_{CM} = \frac{g\sin\theta}{1 + \frac{I_{CM}}{MR^2}}

Minimum Static Friction: —

\mu_{s,min} = \frac{\tan\theta}{1 + \frac{MR^2}{I_{CM}}}

Work by Static Friction (Pure Rolling): — Zero

**Moment of Inertia (

I_{CM}

):**

* Ring/Hollow Cylinder: $MR^2$ * Disc/Solid Cylinder: $\frac{1}{2}MR^2$ * Solid Sphere: $\frac{2}{5}MR^2$ * Hollow Sphere: $\frac{2}{3}MR^2$

To remember the order of objects rolling down an incline (fastest to slowest): Solid Sphere, Solid Cylinder, Hollow Sphere, Hollow Cylinder/Ring.

Think: Super Speedy Cars Have Smooth Heels. (Solid Sphere, Solid Cylinder, Hollow Sphere, Hollow Cylinder/Ring). This mnemonic helps recall the order based on their $I_{CM}/MR^2$ values (smallest to largest).