Mastering Pure Rolling Motion for JEE Physics

Pure rolling motion is a fundamental concept in rotational mechanics, crucial for excelling in competitive exams like JEE. It describes a special case where an object rolls without slipping. Understanding its nuances, including the relationship between translational and rotational motion, energy considerations, and angular momentum, is key.

What is Pure Rolling Motion?

Pure rolling motion occurs when an object (like a wheel or cylinder) rolls along a surface such that the point of contact between the object and the surface is instantaneously at rest relative to the surface. This means there is no slipping or sliding at the point of contact.

No slipping at the point of contact.

In pure rolling, the velocity of the point on the rolling object that touches the surface is zero relative to the surface. This is the defining characteristic.

Mathematically, if $v_{cm}$ is the velocity of the center of mass of the rolling object and $\omega$ is its angular velocity, and $R$ is its radius, then for pure rolling motion on a horizontal surface, the condition is $v_{cm} = R\omega$ . This relationship connects the object's translational motion to its rotational motion.

Key Equations and Relationships

The condition $v_{cm} = R\omega$ is central to solving problems involving pure rolling. Let's explore its implications and related equations.

Parameter	Pure Rolling (Horizontal)	Pure Rolling (Inclined Plane)
Translational Velocity ( $v_{cm}$ )	$v_{cm} = R\omega$	$v_{cm} = R\omega$
Acceleration ( $a_{cm}$ )	$a_{cm} = R\alpha$	$a_{cm} = R\alpha$
Velocity of Top Point	$2v_{cm}$	$2v_{cm}$
Velocity of Bottom Point	$0$	$0$

Here, $\alpha$ is the angular acceleration. The acceleration of the center of mass ( $a_{cm}$ ) is related to the net force, and the angular acceleration ( $\alpha$ ) is related to the net torque. For an object rolling down an inclined plane without slipping, the acceleration is given by $a_{cm} = \frac{g \sin \theta}{1 + \frac{I}{mR^2}}$ , where $\theta$ is the angle of inclination, $I$ is the moment of inertia, and $m$ is the mass.

Energy in Pure Rolling Motion

An object undergoing pure rolling motion possesses both translational kinetic energy and rotational kinetic energy. The total kinetic energy ( $K_{total}$ ) is the sum of these two.

The total kinetic energy of a body in pure rolling motion is the sum of its translational kinetic energy and rotational kinetic energy. Translational kinetic energy is given by $\frac{1}{2}mv_{cm}^2$ , where $m$ is the mass and $v_{cm}$ is the velocity of the center of mass. Rotational kinetic energy is given by $\frac{1}{2}I\omega^2$ , where $I$ is the moment of inertia about the center of mass and $\omega$ is the angular velocity. Combining these, $K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$ . Using the condition for pure rolling, $v_{cm} = R\omega$ , we can express the total kinetic energy solely in terms of $v_{cm}$ or $\omega$ . Substituting $\omega = v_{cm}/R$ , we get $K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I(\frac{v_{cm}}{R})^2 = \frac{1}{2}v_{cm}^2(m + \frac{I}{R^2})$ . This form is particularly useful for applying the work-energy theorem.

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Remember that the point of contact is instantaneously at rest. This is crucial for understanding friction's role (or lack thereof) in maintaining pure rolling.

Angular Momentum in Pure Rolling

The angular momentum of a body in pure rolling motion can be calculated about any point. A convenient point is often the center of mass or the instantaneous point of contact.

The angular momentum ( $L$ ) about the center of mass is $L_{cm} = I_{cm}\omega$ . The total angular momentum about a fixed point (e.g., the ground directly below the center of mass) is the sum of the angular momentum of the center of mass about that point and the angular momentum about the center of mass: $L = mv_{cm}R + I_{cm}\omega$ . For pure rolling, $v_{cm} = R\omega$ , so $L = m(R\omega)R + I_{cm}\omega = (mR^2 + I_{cm})\omega$ . The term $mR^2$ is the moment of inertia of a point mass $m$ at a distance $R$ from the axis, often referred to as the moment of inertia of the center of mass about the point of contact. Thus, the total angular momentum can be seen as $L = I_{total}\omega$ , where $I_{total} = I_{cm} + mR^2$ is the moment of inertia about the instantaneous point of contact.

Common Scenarios and Problem-Solving Strategies

Problems involving pure rolling often require combining concepts from linear motion, rotational motion, energy conservation, and angular momentum conservation. Key strategies include:

Identify the condition for pure rolling: Always ensure $v_{cm} = R\omega$ is applied correctly.
Free-body diagrams: Draw accurate FBDs to identify forces and torques.
Newton's Second Law (Linear): $\sum F_{ext} = ma_{cm}$
Newton's Second Law (Rotational): $\sum \tau_{ext} = I\alpha$
Work-Energy Theorem: Use $K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I\omega^2$ for energy-based solutions.
Angular Momentum Conservation: Apply if no external torque acts about a suitable point.

What is the velocity of the point of contact of a wheel undergoing pure rolling motion relative to the surface?

Zero.

If a wheel rolls down an inclined plane without slipping, does static friction do work?

No, because the point of contact is instantaneously at rest, so there is no displacement at the point of contact where friction acts.

Example Application: Rolling Down an Inclined Plane

Consider a solid cylinder of mass $m$ and radius $R$ rolling down an inclined plane of angle $\theta$ . The moment of inertia of a solid cylinder about its axis is $I = \frac{1}{2}mR^2$ . Using the equations of motion:

Linear motion: $mg\sin\theta - f_s = ma_{cm}$ Rotational motion: $f_s R = I\alpha = (\frac{1}{2}mR^2)\alpha$ Pure rolling condition: $a_{cm} = R\alpha$

Substituting $\alpha = a_{cm}/R$ into the torque equation gives $f_s R = (\frac{1}{2}mR^2)(a_{cm}/R) \implies f_s = \frac{1}{2}ma_{cm}$ .

Substituting $f_s$ into the linear motion equation: $mg\sin\theta - \frac{1}{2}ma_{cm} = ma_{cm}$ .

Solving for $a_{cm}$ : $mg\sin\theta = \frac{3}{2}ma_{cm} \implies a_{cm} = \frac{2}{3}g\sin\theta$ .

This shows that the acceleration is less than $g\sin\theta$ (which would be the case if it were sliding without friction), due to the energy transfer to rotational kinetic energy.

Key Takeaways for JEE

Focus on the condition $v_{cm} = R\omega$ and its implications for velocity and acceleration. Understand how to calculate total kinetic energy and angular momentum. Practice problems involving different shapes (spheres, hollow cylinders, rings) rolling down inclined planes and over obstacles.

Learning Resources

Pure Rolling Motion - Physics Classroom(documentation)

Provides a clear, step-by-step explanation of pure rolling motion, including the fundamental relationships and conditions.

Rotational Motion: Pure Rolling - Khan Academy(video)

A video tutorial explaining the concept of pure rolling motion, its conditions, and how it relates translational and rotational motion.

Pure Rolling Motion - Concepts and Problems(blog)

This resource covers the core concepts of pure rolling motion and provides examples of problems commonly encountered in competitive exams.

Understanding Pure Rolling Motion - Physics Stack Exchange(wikipedia)

A forum discussion that delves into the nuances of pure rolling motion, addressing common questions and misconceptions.

Rolling Motion - Wikipedia(wikipedia)

Provides a comprehensive overview of rolling motion, including pure rolling, slipping, and related physics principles.

Pure Rolling Motion - Toppr(blog)

Explains the concept of pure rolling motion with a focus on its application in JEE Physics, including formulas and problem-solving tips.

Kinetic Energy of Rolling Objects - Physics LibreTexts(documentation)

Details the kinetic energy associated with rolling objects, breaking down translational and rotational components.

Angular Momentum of Rolling Objects - Physics LibreTexts(documentation)

Explains the concept of angular momentum and how it applies to objects in rotational and rolling motion.

JEE Physics: Rotational Motion - Pure Rolling Motion(video)

A detailed video lecture specifically tailored for JEE aspirants, covering pure rolling motion with examples.

Problems on Pure Rolling Motion - Physics Galore(blog)

Offers a collection of solved problems on pure rolling motion, providing practical application of the learned concepts.