Rotational Motion Formulas For JEE 2026
Rotational Motion is one of the most important and frequently tested topics in JEE Physics, with around 2–4 questions commonly appearing in both JEE Main and JEE Advanced. This chapter introduces key concepts such as angular kinematics, torque, moment of inertia, angular momentum and its conservation, rotational kinetic energy, and rolling motion. Since many problems combine multiple concepts from mechanics, a strong understanding of these formulas and principles is essential for solving JEE-level numerical questions accurately.
To master this chapter, students should focus on understanding the relationship between translational and rotational quantities, as this analogy forms the foundation of many rotational motion problems. Regular revision of important formulas can significantly improve speed and accuracy during practice and exams. Using a well-organized JEE Physics Formula PDF can be extremely helpful for quick revision, as it allows students to review key equations and concepts in one place. Referring to a JEE Physics Formula PDF during preparation also helps in strengthening problem-solving skills and ensuring better performance in competitive exams.
What Is Rotational Motion?
So far in physics, you have mostly studied objects moving in straight lines (translational motion). But many real-world objects spin or rotate — a ceiling fan, a wheel, the Earth itself. When a rigid body turns about a fixed axis, every particle in it traces a circle. This is called rotational motion.
In translational motion we use displacement, velocity, and acceleration. Rotational motion has exact parallels: angular displacement, angular velocity, and angular acceleration. Once you see this connection, every rotational formula looks just like its translational counterpart.
Rigid Body: An idealized object where the distance between any two particles remains constant, no matter what forces act on it.
Axis of Rotation: The fixed straight line about which a body rotates. Every particle of the body moves in a circle whose centre lies on this axis.
Angular Kinematics Formulas
Just as we describe straight-line motion with $$s$$, $$v$$, and $$a$$, we describe rotation with three analogous quantities measured in radians.
Angular Displacement ($$\theta$$): The angle (in radians) through which a body has rotated about its axis. One full revolution $$= 2\pi$$ radians $$= 360^\circ$$.
Angular Velocity ($$\omega$$): The rate of change of angular displacement: how fast the object is spinning. Unit: rad/s.
Angular Acceleration ($$\alpha$$): The rate of change of angular velocity: how quickly the spinning speed changes. Unit: rad/s$$^2$$.
Definitions of Angular Quantities
$$$\omega = \frac{d\theta}{dt} \qquad\qquad \alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}$$$where $$\theta$$ = angular displacement (rad), $$\omega$$ = angular velocity (rad/s), $$\alpha$$ = angular acceleration (rad/s$$^2$$).
If the angular acceleration $$\alpha$$ is constant, the equations of rotational kinematics are identical in form to the straight-line SUVAT equations:
Equations of Rotational Kinematics (constant $$\alpha$$)
- $$\omega = \omega_0 + \alpha t$$
- $$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$$
- $$\omega^2 = \omega_0^2 + 2\alpha\theta$$
- $$\theta = \frac{1}{2}(\omega_0 + \omega)\,t$$
where $$\omega_0$$ = initial angular velocity, $$\omega$$ = final angular velocity, $$\theta$$ = angular displacement, $$t$$ = time.
Relation Between Linear and Angular Quantities
- Arc length: $$s = r\theta$$
- Linear speed: $$v = r\omega$$
- Tangential acceleration: $$a_t = r\alpha$$
- Centripetal acceleration: $$a_c = r\omega^2 = \dfrac{v^2}{r}$$
where $$r$$ = distance of the particle from the axis of rotation.
Worked Example
A wheel starts from rest and accelerates uniformly at $$2$$ rad/s$$^2$$. Find the angular velocity and number of revolutions after 5 seconds.
Given: $$\omega_0 = 0$$, $$\alpha = 2$$ rad/s$$^2$$, $$t = 5$$ s
$$\omega = \omega_0 + \alpha t = 0 + 2 \times 5 = 10$$ rad/s
$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}(2)(25) = 25$$ rad
Number of revolutions $$= \dfrac{\theta}{2\pi} = \dfrac{25}{2\pi} \approx 3.98$$ revolutions.
Tip: The rotational kinematics equations are exactly like $$v = u + at$$, $$s = ut + \frac{1}{2}at^2$$, and $$v^2 = u^2 + 2as$$, with the substitutions $$s \to \theta$$, $$u \to \omega_0$$, $$v \to \omega$$, $$a \to \alpha$$.
Moment of Inertia Formulas and Table
In translational motion, mass measures how hard it is to change an object's velocity. In rotational motion, the analogous quantity is the moment of inertia ($$I$$). It depends not only on the mass but also on how that mass is distributed relative to the axis of rotation. Mass far from the axis contributes more to the moment of inertia.
Moment of Inertia ($$I$$): The rotational analogue of mass. For a system of particles: $$I = \sum m_i r_i^2$$, where $$r_i$$ is the perpendicular distance of the $$i$$-th particle from the axis. Unit: kg·m$$^2$$.
$$$I = \sum m_i r_i^2 \qquad \text{(discrete system)} \qquad I = \int r^2 \, dm \qquad \text{(continuous body)}$$$
Moments of Inertia of Common Bodies
| Body | Axis | $$I$$ |
|---|---|---|
| Point mass $$m$$ at distance $$r$$ | Through centre | $$mr^2$$ |
| Thin ring (mass $$M$$, radius $$R$$) | Through centre, perpendicular to plane | $$MR^2$$ |
| Thin ring | Any diameter | $$\frac{1}{2}MR^2$$ |
| Circular disc (mass $$M$$, radius $$R$$) | Through centre, perpendicular to plane | $$\frac{1}{2}MR^2$$ |
| Circular disc | Any diameter | $$\frac{1}{4}MR^2$$ |
| Solid sphere (mass $$M$$, radius $$R$$) | Any diameter | $$\frac{2}{5}MR^2$$ |
| Hollow sphere (thin shell) | Any diameter | $$\frac{2}{3}MR^2$$ |
| Thin rod (mass $$M$$, length $$L$$) | Perpendicular to rod, through centre | $$\frac{1}{12}ML^2$$ |
| Thin rod | Perpendicular to rod, through one end | $$\frac{1}{3}ML^2$$ |
| Solid cylinder (mass $$M$$, radius $$R$$) | Along the axis | $$\frac{1}{2}MR^2$$ |
| Hollow cylinder ($$R_1$$, $$R_2$$) | Along the axis | $$\frac{1}{2}M(R_1^2 + R_2^2)$$ |
Note: The moment of inertia depends on the chosen axis. The same object has different $$I$$ values about different axes. Always check which axis the problem specifies.
Worked Example
Four point masses, each of 2 kg, are placed at the corners of a square of side 1 m. Find the moment of inertia about one side of the square.
Choose one side as the axis. Two masses lie on the axis ($$r = 0$$), and two masses are at perpendicular distance $$r = 1$$ m.
$$I = 2(m \times 0^2) + 2(m \times 1^2) = 0 + 2(2)(1) = 4 \text{ kg·m}^2$$
Parallel Axis Theorem
If you know the moment of inertia about an axis through the centre of mass, you can find it about any parallel axis using this theorem.
Parallel Axis Theorem
$$$I = I_{cm} + Md^2$$$where $$I_{cm}$$ = moment of inertia about an axis through the centre of mass, $$M$$ = total mass, $$d$$ = perpendicular distance between the two parallel axes.
Worked Example
Find the moment of inertia of a uniform rod of mass $$M$$ and length $$L$$ about an axis perpendicular to the rod through one end.
We know $$I_{cm} = \frac{1}{12}ML^2$$ (about centre). The end is at distance $$d = L/2$$ from the centre.
$$I_{\text{end}} = I_{cm} + Md^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \frac{1}{3}ML^2$$ ✓
Perpendicular Axis Theorem
This theorem applies only to flat (planar/laminar) bodies. If you know the moments of inertia about two perpendicular axes lying in the plane of the body, you can find the moment about the axis perpendicular to the plane.
Perpendicular Axis Theorem (Planar Bodies Only)
$$$I_z = I_x + I_y$$$where $$x$$ and $$y$$ are two mutually perpendicular axes in the plane of the body passing through the same point, and $$z$$ is the axis perpendicular to the plane through that point.
Worked Example
The moment of inertia of a disc about its axis (perpendicular to plane) is $$\frac{1}{2}MR^2$$. Find $$I$$ about a diameter.
By symmetry, $$I_x = I_y$$ (any two perpendicular diameters are equivalent).
Using perpendicular axis theorem: $$I_z = I_x + I_y = 2I_x$$
$$\frac{1}{2}MR^2 = 2I_x \Rightarrow I_x = \frac{1}{4}MR^2$$
Tip: The perpendicular axis theorem works only for 2D (laminar) bodies like rings, discs, rectangular plates, etc. It does not apply to 3D objects like spheres or cylinders. JEE often sets traps here.
Torque Formulas
Just as force causes translational acceleration, torque causes angular acceleration. Torque is the rotational equivalent of force. It depends not only on the magnitude of the force but also on where and at what angle the force is applied relative to the axis.
Torque ($$\boldsymbol{\tau}$$): The turning effect of a force about an axis. It is a vector quantity.
Torque
$$$\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F} \qquad \Rightarrow \qquad |\tau| = rF\sin\theta$$$where $$r$$ = distance from the axis to the point of application of force, $$F$$ = magnitude of force, $$\theta$$ = angle between $$\mathbf{r}$$ and $$\mathbf{F}$$.
Alternatively: $$\tau = F \times d$$, where $$d = r\sin\theta$$ is the perpendicular distance from the axis to the line of action of force (called the moment arm).
Unit: N·m. Dimensions: $$[ML^2T^{-2}]$$
Note: Torque has the same dimensions as work/energy ($$[ML^2T^{-2}]$$), but they are physically different quantities. Torque is a vector, while work is a scalar.
Newton's Second Law for Rotation
Just as $$F = ma$$ governs translational motion, a similar law governs rotation:
Newton's Second Law for Rotation
$$$\tau_{\text{net}} = I\alpha$$$where $$\tau_{\text{net}}$$ = net torque about the axis, $$I$$ = moment of inertia about the same axis, $$\alpha$$ = angular acceleration.
Worked Example
A disc of mass 5 kg and radius 0.2 m is free to rotate about its central axis. A tangential force of 10 N is applied at the rim. Find the angular acceleration.
Torque: $$\tau = F \times R = 10 \times 0.2 = 2$$ N·m
Moment of inertia of disc: $$I = \frac{1}{2}MR^2 = \frac{1}{2}(5)(0.04) = 0.1$$ kg·m$$^2$$
$$\alpha = \frac{\tau}{I} = \frac{2}{0.1} = 20$$ rad/s$$^2$$
Angular Momentum and Conservation Formulas
Angular momentum is the rotational analogue of linear momentum. Just as a heavy, fast-moving object is hard to stop, a spinning object with large angular momentum is hard to stop.
Angular Momentum ($$L$$): The rotational equivalent of linear momentum. For a rigid body rotating about a fixed axis: $$L = I\omega$$. Unit: kg·m$$^2$$/s.
Angular Momentum
For a particle: $$\boldsymbol{L} = \mathbf{r} \times \mathbf{p} = \mathbf{r} \times m\mathbf{v} \qquad \Rightarrow \qquad L = mvr\sin\theta$$
For a rigid body about a fixed axis:
$$$L = I\omega$$$Relation to torque (Newton's second law form):
$$$\tau_{\text{net}} = \frac{dL}{dt}$$$Conservation of Angular Momentum
This is one of the most powerful and frequently tested principles in JEE. If no external torque acts on a system, its total angular momentum remains constant.
Conservation of Angular Momentum
If $$\tau_{\text{ext}} = 0$$:
$$$L = I\omega = \text{constant} \qquad \Rightarrow \qquad I_1\omega_1 = I_2\omega_2$$$When moment of inertia decreases, angular velocity increases (and vice versa).
This is why an ice skater spins faster when she pulls her arms in (decreasing $$I$$), and spins slower when she extends them (increasing $$I$$).
Worked Example
A disc of moment of inertia 4 kg·m$$^2$$ is spinning at 10 rad/s. A ring of moment of inertia 2 kg·m$$^2$$ is dropped coaxially onto it. Find the final angular velocity.
No external torque acts on the system, so angular momentum is conserved.
$$I_1\omega_1 = (I_1 + I_2)\omega_f$$
$$4 \times 10 = (4 + 2)\omega_f$$
$$\omega_f = \frac{40}{6} = \frac{20}{3} \approx 6.67$$ rad/s
Tip: Conservation of angular momentum is tested very often in JEE. Common scenarios: collapsing/expanding rotating systems, a person walking on a rotating platform, a bullet hitting a rotating rod.
Rotational Kinetic Energy Formula
A spinning object has kinetic energy due to its rotation, just as a moving object has kinetic energy due to its translation.
Rotational Kinetic Energy
$$$KE_{\text{rot}} = \frac{1}{2}I\omega^2$$$where $$I$$ = moment of inertia, $$\omega$$ = angular velocity.
This is analogous to $$KE_{\text{trans}} = \frac{1}{2}mv^2$$.
The work-energy theorem also has a rotational version:
$$$W = \tau \cdot \theta = \frac{1}{2}I\omega_f^2 - \frac{1}{2}I\omega_i^2$$$
And rotational power:
$$$P = \tau\omega$$$
Worked Example
A flywheel of moment of inertia 0.5 kg·m$$^2$$ is rotating at 600 rpm. Find its rotational kinetic energy.
First convert rpm to rad/s: $$\omega = 600 \times \frac{2\pi}{60} = 20\pi$$ rad/s
$$KE = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.5)(20\pi)^2 = \frac{1}{2}(0.5)(400\pi^2) = 100\pi^2 \approx 987$$ J
Rolling Motion Formulas
When a wheel or ball moves along a surface without slipping, it undergoes rolling motion — a combination of translation of its centre of mass and rotation about its centre of mass. This is the most common type of motion for wheels, balls, and cylinders.
Rolling Without Slipping: The condition where the point of contact with the ground is momentarily at rest. The velocity of the centre of mass equals $$R\omega$$: $$v_{cm} = R\omega$$.
Kinetic Energy in Rolling Motion
Total KE = Translational KE + Rotational KE
$$$KE_{\text{total}} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$$$Using $$v_{cm} = R\omega$$ and $$I_{cm} = kMR^2$$ (where $$k$$ is a geometric constant):
$$$KE_{\text{total}} = \frac{1}{2}Mv_{cm}^2\left(1 + k\right)$$$where $$k = I_{cm}/(MR^2)$$.
| Body | $$k = I_{cm}/(MR^2)$$ | Total KE |
|---|---|---|
| Ring / Hollow cylinder | 1 | $$Mv_{cm}^2$$ |
| Disc / Solid cylinder | $$1/2$$ | $$\frac{3}{4}Mv_{cm}^2$$ |
| Solid sphere | $$2/5$$ | $$\frac{7}{10}Mv_{cm}^2$$ |
| Hollow sphere | $$2/3$$ | $$\frac{5}{6}Mv_{cm}^2$$ |
Rolling Down an Inclined Plane
When a body rolls without slipping down an incline of angle $$\theta$$ and height $$h$$, we can use energy conservation to find its speed at the bottom.
Rolling Down an Incline
Acceleration down the incline:
$$$a = \frac{g\sin\theta}{1 + k}$$$Velocity at the bottom (starting from rest, height $$h$$):
$$$v = \sqrt{\frac{2gh}{1 + k}}$$$Time to reach the bottom (length $$L$$ of incline):
$$$t = \sqrt{\frac{2L(1+k)}{g\sin\theta}}$$$where $$k = I_{cm}/(MR^2)$$.
Note: For rolling down an incline: (1) The body with the smallest $$k$$ reaches the bottom first and has the greatest velocity. (2) A solid sphere ($$k = 2/5$$) always beats a disc ($$k = 1/2$$), which beats a ring ($$k = 1$$). (3) The result does not depend on mass or radius — only on $$k$$.
Worked Example
A solid sphere and a hollow sphere are released simultaneously from the top of an incline of height 2 m. Find the velocity of each at the bottom.
For solid sphere ($$k = 2/5$$): $$v = \sqrt{\frac{2 \times 9.8 \times 2}{1 + 2/5}} = \sqrt{\frac{39.2}{1.4}} = \sqrt{28} \approx 5.29$$ m/s
For hollow sphere ($$k = 2/3$$): $$v = \sqrt{\frac{2 \times 9.8 \times 2}{1 + 2/3}} = \sqrt{\frac{39.2}{5/3}} = \sqrt{23.52} \approx 4.85$$ m/s
The solid sphere is faster and reaches the bottom first.
Tip: In JEE, when asked "which body reaches the bottom first?" the answer is always the one with the lowest $$k$$. The order is: solid sphere, solid cylinder/disc, hollow sphere, hollow cylinder/ring.
Analogy: Translation vs. Rotation
One of the best ways to master rotational motion is to remember the one-to-one correspondence between translational and rotational quantities:
Complete Translation–Rotation Analogy
| Translational | Rotational | Relation |
|---|---|---|
| Displacement $$s$$ | Angular displacement $$\theta$$ | $$s = r\theta$$ |
| Velocity $$v$$ | Angular velocity $$\omega$$ | $$v = r\omega$$ |
| Acceleration $$a$$ | Angular acceleration $$\alpha$$ | $$a_t = r\alpha$$ |
| Mass $$m$$ | Moment of inertia $$I$$ | $$I = \sum mr^2$$ |
| Force $$F$$ | Torque $$\tau$$ | $$\tau = rF\sin\theta$$ |
| Momentum $$p = mv$$ | Angular momentum $$L = I\omega$$ | $$L = r \times p$$ |
| $$F = ma$$ | $$\tau = I\alpha$$ | — |
| $$KE = \frac{1}{2}mv^2$$ | $$KE = \frac{1}{2}I\omega^2$$ | — |
| $$W = Fs$$ | $$W = \tau\theta$$ | — |
| $$P = Fv$$ | $$P = \tau\omega$$ | — |
Tip: JEE problems on rotational motion often combine rotation with translation (e.g., rolling, pulleys with massive wheels). Always identify whether to use $$F = ma$$ for translation, $$\tau = I\alpha$$ for rotation, or both together with the rolling constraint $$a = R\alpha$$.
Rotational Motion Formulas For JEE 2026: Conclusion
Rotational Motion is a fundamental chapter in JEE Physics and plays an important role in both conceptual understanding and problem-solving. Topics such as angular kinematics, torque, moment of inertia, angular momentum, and rolling motion form the foundation for many advanced mechanics problems. Mastering the Rotational Motion Formulas For JEE 2026 helps students solve complex numerical questions efficiently and improves accuracy during the exam.
Regular practice and quick revision of these formulas are essential for success in JEE preparation. Using a structured JEE Physics Formula PDF can help students revise important equations in a short time and strengthen their understanding of key concepts. By consistently reviewing these formulas and applying them in practice problems, students can significantly improve their performance in JEE Main and JEE Advanced.
