Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Systems of Particles and Rotational Motion for CBSE Class 11 Physics, bridging linear mechanics into the rotational world.
Key Concepts
Rotational Mechanics
The physics of spinning and rolling
1
Center of Mass (Discrete Particles)
Formula
The point where the entire mass of a system appears to be concentrated. If no external force acts on the system, the velocity of the center of mass remains absolutely constant.
$$\vec{R}_{\text{cm}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + \dots + m_n\vec{r}_n}{m_1 + m_2 + \dots + m_n} = \frac{\sum m_i \vec{r}_i}{M}$$
2
Torque (Moment of Force)
Formula
The rotational analogue of force. It measures the turning effect of a force about an axis. It is defined as the cross product of the position vector ($\vec{r}$) and the force vector ($\vec{F}$).
$$\vec{\tau} = \vec{r} \times \vec{F} \quad | \quad |\vec{\tau}| = rF \sin\theta$$
3
Angular Momentum
Formula
The rotational analogue of linear momentum. It is the moment of linear momentum about an axis. Also related to Moment of Inertia ($I$) and angular velocity ($\omega$).
$$\vec{L} = \vec{r} \times \vec{p} \quad | \quad L = I\omega$$
4
Newton’s Second Law for Rotation
Formula
The rate of change of angular momentum equals the net external torque applied to the system. If $I$ is constant, Torque equals Moment of Inertia times angular acceleration ($\alpha$).
$$\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt} = I\vec{\alpha}$$
5
Conservation of Angular Momentum
Formula
If the net external torque acting on a system is zero, the total angular momentum of the system remains perfectly conserved.
$$\text{If } \vec{\tau}_{\text{ext}} = 0 \implies \vec{L} = \text{Constant} \implies I_1\omega_1 = I_2\omega_2$$
6
Moment of Inertia ($I$) & Radius of Gyration ($k$)
Formula
The “rotational mass” or rotational inertia of a body. It depends on the mass and how that mass is distributed relative to the axis of rotation. $k$ is the equivalent distance where all mass could be concentrated to give the same $I$.
$$I = \sum m_i r_i^2 \quad | \quad I = \int r^2 dm \quad | \quad k = \sqrt{\frac{I}{M}}$$
7
Theorems of Moment of Inertia
Formula
Perpendicular Axes: For planar (2D) bodies, $I_z = I_x + I_y$.
Parallel Axes: For any body, $I$ about any axis equals $I$ about a parallel axis through the COM plus $Md^2$.
Parallel Axes: For any body, $I$ about any axis equals $I$ about a parallel axis through the COM plus $Md^2$.
$$I_z = I_x + I_y \quad | \quad I = I_{\text{cm}} + Md^2$$
8
Rotational Kinematics
Formula
The rotational equivalents of the Big Three linear equations. Valid ONLY when angular acceleration ($\alpha$) is constant.
$$\omega = \omega_0 + \alpha t$$
$$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$$
$$\omega^2 = \omega_0^2 + 2\alpha\theta$$
9
Rotational Kinetic Energy
Formula
The energy possessed by a body due to its rotation about an axis. Note the perfect analogy to $K = \frac{1}{2}mv^2$.
$$K_R = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$$
10
Pure Rolling Motion
Case Formulas
A combination of translation and rotation where the point of contact with the ground has zero instantaneous velocity. Total Kinetic energy includes both linear and rotational parts.
$$v_{\text{cm}} = \omega R \quad | \quad K_{\text{total}} = \frac{1}{2}M v_{\text{cm}}^2 + \frac{1}{2}I \omega^2 = \frac{1}{2}M v_{\text{cm}}^2 \left(1 + \frac{k^2}{R^2}\right)$$
11
Acceleration of Rolling Body on an Incline
Formula
When a rigid body of mass $M$, radius $R$, and radius of gyration $k$ rolls down a rough inclined plane of angle $\theta$ without slipping.
$$a = \frac{g \sin\theta}{1 + \frac{I}{MR^2}} = \frac{g \sin\theta}{1 + \frac{k^2}{R^2}}$$
Concept Deep Dive
01
Conservation of Angular Momentum
The Ice Skater EffectIf you sit on a spinning chair with dumbbells extended, you have a large Moment of Inertia ($I_1$) and a certain angular velocity ($\omega_1$). Because the bearing is frictionless, external torque $\tau_{\text{ext}} = 0$, meaning Angular Momentum ($L = I\omega$) MUST stay constant.
When you pull your arms in, you drastically decrease your radius, which drops your Moment of Inertia to a much smaller $I_2$. To keep the total product $I \times \omega$ constant, your new angular velocity ($\omega_2$) must massively increase! You spin much faster without any external force pushing you.
When you pull your arms in, you drastically decrease your radius, which drops your Moment of Inertia to a much smaller $I_2$. To keep the total product $I \times \omega$ constant, your new angular velocity ($\omega_2$) must massively increase! You spin much faster without any external force pushing you.
$$I_1 \omega_1 = I_2 \omega_2$$
02
The Magic of Pure Rolling
Why the bottom of a tire is completely stationaryWhen a tire rolls on a road without slipping, it is doing two things at once: Translating forward with velocity $v_{\text{cm}}$ and Rotating with velocity $\omega R$.
At the very top of the tire, translation ($+v_{\text{cm}}$) and rotation ($+\omega R$) point in the SAME direction, making the top move at $2v_{\text{cm}}$.
At the very bottom of the tire (the contact point), translation is forward ($+v_{\text{cm}}$) but the rotation is pushing backward ($-\omega R$). In pure rolling, $v_{\text{cm}} = \omega R$, meaning they perfectly cancel out. The bottom point of your tire has a velocity of exactly zero relative to the road! This is why static friction—not kinetic friction—acts on a purely rolling wheel.
At the very top of the tire, translation ($+v_{\text{cm}}$) and rotation ($+\omega R$) point in the SAME direction, making the top move at $2v_{\text{cm}}$.
At the very bottom of the tire (the contact point), translation is forward ($+v_{\text{cm}}$) but the rotation is pushing backward ($-\omega R$). In pure rolling, $v_{\text{cm}} = \omega R$, meaning they perfectly cancel out. The bottom point of your tire has a velocity of exactly zero relative to the road! This is why static friction—not kinetic friction—acts on a purely rolling wheel.
Compare & Contrast
✗ Linear (Translational) Motion
- Inertia is determined solely by Mass ($m$).
- Motion caused by Force ($F$).
- Momentum is $p = mv$.
- Kinetic Energy is $K = \frac{1}{2}mv^2$.
- Newton’s Law: $F = ma$.
✓ Rotational Motion
- Inertia is determined by Moment of Inertia ($I$).
- Motion caused by Torque ($\tau$).
- Angular Momentum is $L = I\omega$.
- Kinetic Energy is $K_R = \frac{1}{2}I\omega^2$.
- Newton’s Law: $\tau = I\alpha$.
Remember
Moment of Inertia ($I$) is NOT a fixed constant like Mass ($m$). A single solid object has an infinite number of Moments of Inertia depending on exactly where you place the axis of rotation!
Common Mistakes to Avoid
Mistake 1
Misusing the Parallel Axis Theorem: The formula is $I = I_{\text{cm}} + Md^2$. Students often try to apply this between ANY two parallel axes. This is strictly wrong! The theorem ONLY works if one of the axes passes exactly through the Center of Mass ($I_{\text{cm}}$).
Mistake 2
Cross Product Order for Torque: Torque is mathematically defined as $\vec{\tau} = \vec{r} \times \vec{F}$. Many students write it as $\vec{F} \times \vec{r}$. Because cross products are anti-commutative ($\vec{A} \times \vec{B} = -\vec{B} \times \vec{A}$), flipping the order will give you the exact opposite direction for the torque vector!
Mistake 3
Confusing Constant Angular Velocity with Equilibrium: If a wheel is spinning at a constant $100 \text{ rad/s}$, its angular acceleration is zero ($\alpha = 0$). By $\tau = I\alpha$, the net torque is zero. The system is in rotational equilibrium, even though it is moving very fast!
Exam Tips
Tip 1
The Shape Factor ($c = k^2/R^2$): Memorize these constants for Rolling Race problems! Ring/Hollow Cylinder: $1$. Solid Cylinder/Disk: $0.5$. Hollow Sphere: $2/3$. Solid Sphere: $0.4$. The shape with the lowest fraction (Solid Sphere) has the least rotational inertia and will ALWAYS win a race down an incline!
Tip 2
If a bomb explodes mid-air, internal chemical forces break it apart. Since external force ($\vec{F}_{\text{ext}}$) is zero, the Center of Mass of the fragments will continue to follow the exact same parabolic trajectory it was on before the explosion!
Expected Exam Questions
SQ
Board Pattern Questions
Class 11 · Rotational Motion · CBSE Exam1
Find the torque of a force $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ acting at the point $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$. [2 marks]
Answer
$\vec{\tau} = 14\hat{i} – 38\hat{j} + 16\hat{k}$
📝
Explanation
Torque is calculated using the cross product determinant: $\vec{\tau} = \vec{r} \times \vec{F}$.
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$
$= \hat{i}[(3)(5) – (1)(1)] – \hat{j}[(7)(5) – (1)(-3)] + \hat{k}[(7)(1) – (3)(-3)]$
$= \hat{i}[15 – 1] – \hat{j}[35 + 3] + \hat{k}[7 + 9]$
$= 14\hat{i} – 38\hat{j} + 16\hat{k}$ N·m.
2
Three identical spheres, each of mass $M$ and radius $R$, are placed at the corners of an equilateral triangle of side $2R$ (touching each other). Find the moment of inertia of the system about an axis passing through the center of one sphere and perpendicular to the plane of the triangle. [3 marks]
Answer
$\frac{46}{5} MR^2$
📝
Explanation
Let the axis pass through Sphere 1.
1. MOI of Sphere 1 about its own diameter (passing through COM): $I_1 = \frac{2}{5}MR^2$.
2. MOI of Sphere 2: Its COM is at distance $d = 2R$ from the axis. Using Parallel Axis Theorem: $I_2 = I_{\text{cm}} + Md^2 = \frac{2}{5}MR^2 + M(2R)^2 = \frac{2}{5}MR^2 + 4MR^2 = \frac{22}{5}MR^2$.
3. MOI of Sphere 3 is identical to Sphere 2: $I_3 = \frac{22}{5}MR^2$.
4. Total MOI of the system $I_{\text{total}} = I_1 + I_2 + I_3 = \frac{2}{5}MR^2 + \frac{22}{5}MR^2 + \frac{22}{5}MR^2 = \frac{46}{5}MR^2$.
3
A child stands at the center of a turntable with his two arms outstretched. The turntable is set rotating with an angular speed of $40 \text{ rev/min}$. If he folds his hands back, reducing his moment of inertia to $2/5$ of the initial value, what will be the new angular speed? [3 marks]
Answer
$100 \text{ rev/min}$
📝
Explanation
Since no external torque acts on the child-turntable system, Angular Momentum is conserved.
$L_1 = L_2 \implies I_1 \omega_1 = I_2 \omega_2$
Given: Initial MOI = $I_1$. Initial angular speed $\omega_1 = 40 \text{ rpm}$.
Final MOI $I_2 = \frac{2}{5}I_1$.
$I_1 (40) = \left(\frac{2}{5} I_1\right) \omega_2$
$40 = \frac{2}{5} \omega_2 \implies \omega_2 = 40 \times \frac{5}{2} = 20 \times 5 = 100 \text{ rev/min}$.
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