This sheet collects the Systems of Particles and Rotational Motion Class 11 formulas from NCERT Physics Part I, Chapter 6 — centre of mass, motion of the centre of mass, linear momentum of a system, vector product, angular velocity, torque and angular momentum, moment of inertia, and the kinematics and dynamics of rotation about a fixed axis.
Every formula is grouped by the textbook’s own sub-topic, with the meaning and SI unit of each symbol.
Use this page as a lookup sheet while revising: read the formula group you need, confirm what each symbol means, check the when-to-use guidance, and test yourself on the worked examples. For the detailed derivations and explanations, the Class 11 Physics formula index gathers this chapter’s material alongside the rest of the class.
Formulas at a Glance
The complete set of formulas on this sheet, in one table. Conditions for each formula are given in the when-to-use section.
| Purpose (what you are finding) | Formula |
|---|---|
| Centre of mass of two particles on a line | \( X = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \) |
| Centre of mass of n particles | \( \mathbf{R} = \frac{\sum m_i\mathbf{r}_i}{M} \) |
| Centre of mass of a continuous body | \( \mathbf{R} = \frac{1}{M}\int\mathbf{r}\,dm \) |
| Velocity of the centre of mass | \( M\mathbf{V} = \sum m_i\mathbf{v}_i \) |
| Acceleration of the centre of mass | \( M\mathbf{A} = \mathbf{F}_{\text{ext}} \) |
| Total linear momentum of a system | \( \mathbf{P} = M\mathbf{V} \) |
| Newton’s second law for a system of particles | \( \frac{d\mathbf{P}}{dt} = \mathbf{F}_{\text{ext}} \) |
| Conservation of linear momentum (when the total external force is zero) | \( \mathbf{P} = \text{constant} \) |
| Magnitude of the vector (cross) product | \( |\mathbf{a}\times\mathbf{b}| = ab\sin\theta \) |
| Cross product in component (determinant) form | \( \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \) |
| Linear speed of a particle of a rotating body | \( v = \omega r \) |
| Vector relation between linear and angular velocity | \( \mathbf{v} = \boldsymbol{\omega}\times\mathbf{r} \) |
| Angular acceleration | \( \alpha = \frac{d\omega}{dt} \) |
| Torque (moment of force) | \( \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \) |
| Magnitude of torque using the lever arm | \( \tau = rF\sin\theta = r_\perp F = rF_\perp \) |
| Angular momentum of a single particle | \( \mathbf{l} = \mathbf{r}\times\mathbf{p} \) |
| Angular momentum of a system of particles | \( \mathbf{L} = \sum \mathbf{r}_i\times\mathbf{p}_i \) |
| Rotational analogue of Newton’s second law | \( \frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{ext}} \) |
| Conservation of angular momentum (when the total external torque is zero) | \( \mathbf{L} = \text{constant} \) |
| Translational equilibrium of a rigid body | \( \sum \mathbf{F}_i = \mathbf{0} \) |
| Rotational equilibrium of a rigid body | \( \sum \boldsymbol{\tau}_i = \mathbf{0} \) |
| Principle of moments for a lever | \( d_1F_1 = d_2F_2 \) |
| Mechanical advantage of a lever | \( \text{M.A.} = \frac{F_1}{F_2} = \frac{d_2}{d_1} \) |
| Moment of inertia of a system of particles | \( I = \sum m_i r_i^2 \) |
| Rotational kinetic energy | \( K = \frac{1}{2}I\omega^2 \) |
| Radius of gyration | \( I = Mk^2 \) |
| Moment of inertia of regular bodies (Table 6.1) | \( I_{\text{ring}} = MR^2,\ I_{\text{disc}} = \frac{MR^2}{2},\ I_{\text{rod}} = \frac{ML^2}{12},\ I_{\text{sphere}} = \frac{2}{5}MR^2 \) |
| Angular velocity after time t (constant angular acceleration) | \( \omega = \omega_0 + \alpha t \) |
| Angular displacement after time t (constant angular acceleration) | \( \theta = \theta_0 + \omega_0t + \frac{1}{2}\alpha t^2 \) |
| Angular velocity–displacement relation (constant angular acceleration) | \( \omega^2 = \omega_0^2 + 2\alpha(\theta – \theta_0) \) |
| Work done by a torque | \( dW = \tau\,d\theta \) |
| Power delivered by a torque | \( P = \tau\omega \) |
| Newton’s second law for rotation about a fixed axis | \( \tau = I\alpha \) |
| Angular momentum for rotation about a fixed axis | \( \mathbf{L}_z = I\omega\hat{\mathbf{k}} \) |
| General law for fixed-axis rotation | \( \frac{d}{dt}(I\omega) = \tau \) |
| Conservation of angular momentum about a fixed axis (when the external torque is zero) | \( I\omega = \text{constant} \) |
All Formulas, Grouped by Topic
Centre of Mass
For two particles on the x-axis, the centre of mass is the mass-weighted mean of their positions (NCERT, p. 96):
\[ X = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \]
For equal masses \( m_1 = m_2 \), this gives \( X = (x_1 + x_2)/2 \), the midpoint. For n particles in space (NCERT, p. 97):
\[ \mathbf{R} = \frac{\sum m_i\mathbf{r}_i}{M}, \qquad M = \sum m_i \]
In coordinates: \( X = \sum m_i x_i / M \), \( Y = \sum m_i y_i / M \), \( Z = \sum m_i z_i / M \). If the origin is placed at the centre of mass, then \( \sum m_i\mathbf{r}_i = 0 \). For a continuous body the sums become integrals:
\[ \mathbf{R} = \frac{1}{M}\int\mathbf{r}\,dm, \qquad X = \frac{1}{M}\int x\,dm \]
By symmetry, the centre of mass of a homogeneous ring, disc, sphere or rod lies at its geometric centre (NCERT, p. 97).
Motion of Centre of Mass
Differentiating \( M\mathbf{R} = \sum m_i\mathbf{r}_i \) once gives the velocity, and twice gives the acceleration of the centre of mass (NCERT, p. 99):
\[ M\mathbf{V} = \sum m_i\mathbf{v}_i, \qquad M\mathbf{A} = \mathbf{F}_{\text{ext}} \]
The second equation states that the centre of mass moves as if the total mass of the system were concentrated there and all external forces were applied at that point. Internal forces cancel in equal-and-opposite pairs, so they never affect the motion of the centre of mass.
Linear Momentum of a System of Particles
The total linear momentum of a system equals the total mass times the velocity of the centre of mass (NCERT, p. 100):
\[ \mathbf{P} = M\mathbf{V} \]
\[ \frac{d\mathbf{P}}{dt} = \mathbf{F}_{\text{ext}} \]
Conservation of linear momentum: if \( \mathbf{F}_{\text{ext}} = 0 \), then \( \mathbf{P} = \text{constant} \). Equivalently, the centre of mass moves with constant velocity.
Vector Product of Two Vectors
The cross product \( \mathbf{c} = \mathbf{a}\times\mathbf{b} \) is a vector perpendicular to the plane of \( \mathbf{a} \) and \( \mathbf{b} \), with magnitude (NCERT, p. 102):
\[ |\mathbf{c}| = ab\sin\theta \]
Its direction is given by the right-hand (or right-handed screw) rule, taking the smaller angle \( \theta \lt 180^\circ \). Key properties:
- Not commutative: \( \mathbf{a}\times\mathbf{b} = -\mathbf{b}\times\mathbf{a} \).
- Unit vectors: \( \hat{\mathbf{i}}\times\hat{\mathbf{j}} = \hat{\mathbf{k}} \), \( \hat{\mathbf{j}}\times\hat{\mathbf{k}} = \hat{\mathbf{i}} \), \( \hat{\mathbf{k}}\times\hat{\mathbf{i}} = \hat{\mathbf{j}} \); reversing the order changes the sign.
Component form, evaluated as a determinant:
\[ \mathbf{a}\times\mathbf{b} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \]
Angular Velocity and Its Relation with Linear Velocity
Every particle of a rigid body rotating about a fixed axis has the same angular velocity \( \omega \), but its linear speed depends on the perpendicular distance from the axis (NCERT, p. 104):
\[ v_i = \omega r_i, \qquad \mathbf{v} = \boldsymbol{\omega}\times\mathbf{r} \]
Angular acceleration is the time rate of change of angular velocity:
\[ \alpha = \frac{d\omega}{dt} \]
Torque and Angular Momentum
Torque (moment of force) about the origin is the cross product of the position vector and the force (NCERT, p. 106):
\[ \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}, \qquad \tau = rF\sin\theta = r_\perp F = rF_\perp \]
Its SI unit is \( \text{N m} \) and its dimensions are \( ML^2T^{-2} \) — the same dimensions as work, but torque is a vector while work is a scalar. The torque vanishes if the line of action of \( \mathbf{F} \) passes through the origin.
Angular momentum of a single particle:
\[ \mathbf{l} = \mathbf{r}\times\mathbf{p}, \qquad l = rp\sin\theta \]
For a system of particles (NCERT, p. 107):
\[ \mathbf{L} = \sum \mathbf{r}_i\times\mathbf{p}_i \]
\[ \frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{ext}} \]
Conservation of angular momentum: if \( \boldsymbol{\tau}_{\text{ext}} = 0 \), then \( \mathbf{L} = \text{constant} \) (NCERT, p. 108).
Equilibrium of a Rigid Body
A rigid body is in mechanical equilibrium when both conditions hold (NCERT, p. 109):
\[ \sum \mathbf{F}_i = \mathbf{0} \quad \text{and} \quad \sum \boldsymbol{\tau}_i = \mathbf{0} \]
These two vector equations contain six scalar conditions; for coplanar forces only three are independent. A couple — equal and opposite forces with different lines of action — has zero net force but non-zero torque, so it produces pure rotation without translation.
For a lever, taking moments about the fulcrum gives the principle of moments (NCERT, p. 111):
\[ d_1F_1 = d_2F_2, \qquad \text{M.A.} = \frac{F_1}{F_2} = \frac{d_2}{d_1} \]
The centre of gravity is the point where the total gravitational torque on the body is zero: \( \sum \mathbf{r}_i\times m_i\mathbf{g} = 0 \). In uniform gravity it coincides with the centre of mass (NCERT, p. 112).
Moment of Inertia
Moment of inertia is the rotational analogue of mass (NCERT, p. 114):
\[ I = \sum m_i r_i^2 \]
Its dimensions are \( ML^2 \) and its SI unit is \( \text{kg m}^2 \). The rotational kinetic energy is:
\[ K = \frac{1}{2}I\omega^2 \]
The radius of gyration \( k \) is defined by \( I = Mk^2 \): it is the distance from the axis at which the whole mass could be concentrated to give the same moment of inertia.
Moments of inertia of regular homogeneous bodies about standard axes (NCERT, p. 115, Table 6.1):
| Body | Axis | Moment of inertia \( I \) |
|---|---|---|
| Thin circular ring, radius R | Perpendicular to plane, at centre | \( MR^2 \) |
| Thin circular ring, radius R | Diameter | \( MR^2/2 \) |
| Thin rod, length L | Perpendicular to rod, at midpoint | \( ML^2/12 \) |
| Circular disc, radius R | Perpendicular to disc, at centre | \( MR^2/2 \) |
| Circular disc, radius R | Diameter | \( MR^2/4 \) |
| Hollow cylinder, radius R | Axis of cylinder | \( MR^2 \) |
| Solid cylinder, radius R | Axis of cylinder | \( MR^2/2 \) |
| Solid sphere, radius R | Diameter | \( 2MR^2/5 \) |
Kinematics of Rotational Motion about a Fixed Axis
For constant angular acceleration, the equations are exactly parallel to the linear ones (NCERT, p. 117):
\[ \omega = \omega_0 + \alpha t \]
\[ \theta = \theta_0 + \omega_0t + \frac{1}{2}\alpha t^2 \]
\[ \omega^2 = \omega_0^2 + 2\alpha(\theta – \theta_0) \]
Dynamics of Rotational Motion about a Fixed Axis
Table 6.2 gives the linear-to-rotational correspondence that makes these formulas easy to remember (NCERT, p. 119):
| Quantity | Linear motion | Rotational motion about a fixed axis |
|---|---|---|
| Displacement | \( x \) | \( \theta \) |
| Velocity | \( v = dx/dt \) | \( \omega = d\theta/dt \) |
| Acceleration | \( a = dv/dt \) | \( \alpha = d\omega/dt \) |
| Mass / inertia | \( M \) | \( I \) |
| Force / torque | \( F = Ma \) | \( \tau = I\alpha \) |
| Work | \( dW = F\,ds \) | \( dW = \tau\,d\theta \) |
| Kinetic energy | \( K = Mv^2/2 \) | \( K = I\omega^2/2 \) |
| Power | \( P = Fv \) | \( P = \tau\omega \) |
| Momentum | \( p = Mv \) | \( L = I\omega \) |
Work, power and Newton’s second law for rotation:
\[ dW = \tau\,d\theta, \qquad P = \tau\omega, \qquad \tau = I\alpha \]
\( \tau = I\alpha \) holds when \( I \) is constant; the general fixed-axis statement is \( \tau = d(I\omega)/dt \) (NCERT, p. 120).
Angular Momentum in Case of Rotation about a Fixed Axis
For fixed-axis rotation, the component of angular momentum along the axis is (NCERT, p. 121):
\[ \mathbf{L}_z = I\omega\hat{\mathbf{k}} \]
For a body symmetric about the axis of rotation, \( \mathbf{L} = \mathbf{L}_z = I\omega\hat{\mathbf{k}} \); for other bodies \( \mathbf{L} \) is not along the axis. The general fixed-axis equation is:
\[ \frac{d}{dt}(I\omega) = \tau \]
Conservation of angular momentum (fixed axis): if \( \tau = 0 \), then \( I\omega = \text{constant} \) — even when \( I \) changes, \( \omega \) adjusts to keep the product constant (NCERT, p. 123).
What Each Symbol Means
| Symbol | What it means | SI unit |
|---|---|---|
| \( m_i \) | Mass of the i-th particle | \( \text{kg} \) |
| \( M = \sum m_i \) | Total mass of the system | \( \text{kg} \) |
| \( \mathbf{r}_i \) | Position vector of the i-th particle from the origin | \( \text{m} \) |
| \( x_i, y_i, z_i \) | Coordinates of the i-th particle | \( \text{m} \) |
| \( \mathbf{R} \) | Position vector of the centre of mass | \( \text{m} \) |
| \( X, Y, Z \) | Coordinates of the centre of mass | \( \text{m} \) |
| \( \mathbf{V} \) | Velocity of the centre of mass | \( \text{m/s} \) |
| \( \mathbf{A} \) | Acceleration of the centre of mass | \( \text{m/s}^2 \) |
| \( \mathbf{v}_i \) | Velocity of the i-th particle | \( \text{m/s} \) |
| \( \mathbf{p} = m\mathbf{v} \) | Linear momentum of a particle | \( \text{kg m/s} \) |
| \( \mathbf{P} \) | Total linear momentum of the system | \( \text{kg m/s} \) |
| \( \mathbf{F}_{\text{ext}} \) | Sum of all external forces on the system | \( \text{N} \) |
| \( \mathbf{a}, \mathbf{b} \) | The two vectors in a cross product | Same as the quantity each vector represents |
| \( \theta \) | Angle between \( \mathbf{a} \) and \( \mathbf{b} \), taken through the smaller angle \( \lt 180^\circ \) | \( \text{rad} \) (dimensionless) |
| \( \theta \) | Angular displacement of a rotating body | \( \text{rad} \) |
| \( \omega \) | Angular velocity (same for every particle of a rigid body) | \( \text{rad/s} \) |
| \( \omega_0 \) | Initial angular velocity, at \( t = 0 \) | \( \text{rad/s} \) |
| \( \alpha \) | Angular acceleration | \( \text{rad/s}^2 \) |
| \( v \) | Linear (tangential) speed of a particle of a rotating body | \( \text{m/s} \) |
| \( r, r_i \) | Perpendicular distance of a particle from the rotation axis | \( \text{m} \) |
| \( r_\perp \) | Perpendicular distance of the line of action of \( \mathbf{F} \) from the origin (lever arm) | \( \text{m} \) |
| \( F_\perp \) | Component of \( \mathbf{F} \) perpendicular to \( \mathbf{r} \) | \( \text{N} \) |
| \( \boldsymbol{\tau} \) | Torque (moment of force), \( \mathbf{r}\times\mathbf{F} \) | \( \text{N m} \) (dimension \( ML^2T^{-2} \)) |
| \( \mathbf{l} \) | Angular momentum of a single particle, \( \mathbf{r}\times\mathbf{p} \) | \( \text{kg m}^2/\text{s} \) |
| \( \mathbf{L} \) | Total angular momentum of a system of particles | \( \text{kg m}^2/\text{s} \) |
| \( I \) | Moment of inertia about the given axis | \( \text{kg m}^2 \) (dimension \( ML^2 \)) |
| \( k \) | Radius of gyration | \( \text{m} \) |
| \( K \) | Rotational kinetic energy | \( \text{J} \) |
| \( W \) | Work done by a torque | \( \text{J} \) |
| \( P \) | Power delivered by a torque | \( \text{W} \) |
| \( t \) | Time | \( \text{s} \) |
| \( d_1, d_2 \) | Load arm and effort arm (distances from the fulcrum) | \( \text{m} \) |
| \( F_1, F_2 \) | Load and effort | \( \text{N} \) |
| \( \mathbf{g} \) | Acceleration due to gravity | \( \text{m/s}^2 \) |
| \( \hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}} \) | Unit vectors along the x-, y- and z-axes | Dimensionless |
When to Use Each Formula
| Formula | Use it when… | Condition that must hold |
|---|---|---|
| \( \mathbf{R} = \sum m_i\mathbf{r}_i / M \) (or the two-particle form) | You need the balance point of a system of discrete particles | Positions of all particles are known, measured from the same origin |
| \( \mathbf{R} = \frac{1}{M}\int\mathbf{r}\,dm \) | The body is continuous, such as a rod, disc or lamina | For homogeneous regular bodies, CM is at the geometric centre by symmetry |
| \( M\mathbf{A} = \mathbf{F}_{\text{ext}} \), \( \mathbf{P} = M\mathbf{V} \) | You want only the translational motion or total momentum of a system | Total mass constant; internal forces cancel and never appear |
| \( \mathbf{P} = \text{constant} \) | Explosions, collisions or decays where the system changes internally | Total external force \( \mathbf{F}_{\text{ext}} = 0 \) |
| \( \mathbf{a}\times\mathbf{b} \), \( ab\sin\theta \) | Defining torque, angular momentum and \( \mathbf{v} = \boldsymbol{\omega}\times\mathbf{r} \) | Order matters: \( \mathbf{a}\times\mathbf{b} = -\mathbf{b}\times\mathbf{a} \); use the smaller angle \( \theta \lt 180^\circ \) |
| \( v = \omega r \) | Converting angular velocity to the linear speed of a point of a rotating body | \( r \) is the perpendicular distance from the axis; \( \omega \) in \( \text{rad/s} \) |
| \( \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} \), \( \tau = r_\perp F \) | The turning effect of a force: doors, levers, torque about a point | Torque is zero if the line of action passes through the point |
| \( \mathbf{L} = \sum\mathbf{r}_i\times\mathbf{p}_i \), \( d\mathbf{L}/dt = \boldsymbol{\tau}_{\text{ext}} \) | The rotational state of a system and how external torque changes it | All moments must be taken about the same origin |
| \( \mathbf{L} = \text{constant} \) | No external torque acts — planets, spinning skaters, rotating wheels | Total external torque \( \boldsymbol{\tau}_{\text{ext}} = 0 \) |
| \( \sum\mathbf{F}_i = \mathbf{0} \), \( \sum\boldsymbol{\tau}_i = \mathbf{0} \) | Static equilibrium problems — ladders, rods on supports, see-saws | Both conditions are needed together for mechanical equilibrium |
| \( d_1F_1 = d_2F_2 \), \( \text{M.A.} = d_2/d_1 \) | Levers, balances and beams pivoted at a fulcrum | Moments are taken about the fulcrum |
| \( I = \sum m_i r_i^2 \), \( K = \frac{1}{2}I\omega^2 \), \( I = Mk^2 \) | Rotational inertia, rotational kinetic energy, radius of gyration | \( I \) depends on the chosen axis; use Table 6.1 values for regular bodies |
| \( \omega = \omega_0 + \alpha t \), \( \theta = \theta_0 + \omega_0t + \frac{1}{2}\alpha t^2 \), \( \omega^2 = \omega_0^2 + 2\alpha(\theta-\theta_0) \) | Rotation with uniform angular acceleration, such as a motor wheel speeding up | \( \alpha \) is constant; \( \omega \) in \( \text{rad/s} \) |
| \( \tau = I\alpha \), \( dW = \tau\,d\theta \), \( P = \tau\omega \) | Angular acceleration, work and power in rotation about a fixed axis | \( \tau = I\alpha \) requires constant \( I \) (rigid body, fixed axis) |
| \( I\omega = \text{constant} \) | Fixed-axis rotation when \( I \) changes — arms in/out, divers, acrobats | No external torque about the axis |
Worked Examples
Example 1: Centre of mass of three particles
Step 1: Three particles — 2.0 kg at (0 m, 0 m), 3.0 kg at (4.0 m, 0 m) and 5.0 kg at (0, 3.0 m).
Total mass \( M = 2.0 + 3.0 + 5.0 = 10.0\ \text{kg} \).
Use \( \mathbf{R} = \sum m_i\mathbf{r}_i / M \), applied to the x- and y-coordinates separately.
All coordinates are measured from the chosen origin.
\[ X = \frac{2.0(0) + 3.0(4.0) + 5.0(0)}{10.0}\ \text{m} = \frac{12.0}{10.0}\ \text{m} = 1.2\ \text{m} \]
\[ Y = \frac{2.0(0) + 3.0(0) + 5.0(3.0)}{10.0}\ \text{m} = \frac{15.0}{10.0}\ \text{m} = 1.5\ \text{m} \]
Final answer: The centre of mass is at \( (1.2\ \text{m},\ 1.5\ \text{m}) \). It lies closer to the 5.0 kg particle because that particle is the heaviest.
Example 2: Conservation of angular momentum — finding the new moment of inertia
Step 1: A skater spins with \( \omega_1 = 2.0\ \text{rad/s} \) and moment of inertia \( I_1 = 6.0\ \text{kg m}^2 \).
Friction is negligible, so no external torque acts about the spin axis and angular momentum is conserved: \( I_1\omega_1 = I_2\omega_2 \).
After pulling her arms in, \( \omega_2 = 6.0\ \text{rad/s} \).
\[ I_2 = \frac{I_1\omega_1}{\omega_2} = \frac{6.0\ \text{kg m}^2 \times 2.0\ \text{rad/s}}{6.0\ \text{rad/s}} = 2.0\ \text{kg m}^2 \]
Final answer: The moment of inertia falls to \( 2.0\ \text{kg m}^2 \). Because \( I\omega \) stays constant, the angular speed triples when \( I \) drops to one-third.
Example 3: Rotational kinematics with constant angular acceleration
Step 1: A wheel starts from rest, so \( \omega_0 = 0 \), and turns with constant \( \alpha = 4.0\ \text{rad/s}^2 \).
Find \( \omega \) after it has turned through \( \theta – \theta_0 = 200\ \text{rad} \).
The relation without time is \( \omega^2 = \omega_0^2 + 2\alpha(\theta – \theta_0) \).
\[ \omega^2 = 0 + 2(4.0\ \text{rad/s}^2)(200\ \text{rad}) = 1600\ (\text{rad/s})^2 \]
\[ \omega = 40\ \text{rad/s} \]
Final answer: The angular speed is \( 40\ \text{rad/s} \), which is about 6.4 revolutions per second.
Common Mistakes to Avoid
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Using the full length of the position vector instead of the perpendicular distance from the axis in \( v = \omega r \) or \( I = \sum m_i r_i^2 \) | In both formulas, \( r \) is the perpendicular distance of the particle from the rotation axis | A particle on the axis must have \( v = 0 \) and contribute zero to \( I \) |
| Reversing the order in a cross product and keeping the same direction | \( \mathbf{a}\times\mathbf{b} = -\mathbf{b}\times\mathbf{a} \); direction comes from the right-hand rule | Recompute with the order swapped — the vector must reverse direction |
| Applying \( \tau = I\alpha \) when \( I \) is changing, such as a skater pulling her arms in | \( \tau = I\alpha \) needs constant \( I \); the general fixed-axis law is \( \tau = d(I\omega)/dt \) | When \( I \) changes and no external torque acts, use \( I_1\omega_1 = I_2\omega_2 \) |
| Treating rpm as if it were already rad/s | Convert first: \( \omega = \frac{2\pi\times\text{rpm}}{60}\ \text{rad/s} \) | Check the final angular speed is in \( \text{rad/s} \), not \( \text{rev/min} \) |
| Confusing moments of inertia of different bodies, especially ring and disc | Ring through centre: \( MR^2 \); disc through centre: \( MR^2/2 \); rod about midpoint: \( ML^2/12 \); solid sphere about diameter: \( 2MR^2/5 \) | Spot-check against Table 6.1 — mass distributed farther from the axis gives larger \( I \) |
| Declaring a body in equilibrium after checking only \( \sum\mathbf{F} = 0 \) | Mechanical equilibrium needs both \( \sum\mathbf{F}_i = \mathbf{0} \) and \( \sum\boldsymbol{\tau}_i = \mathbf{0} \) | A couple has zero net force but still rotates the body, so the torque condition is essential |
Frequently Asked Questions
How do I revise Systems of Particles and Rotational Motion Class 11 formulas quickly?
Start with the at-a-glance table, then use the when-to-use table to practise choosing the right formula for a situation. Do one centre of mass calculation, one conservation of angular momentum problem and one rotational kinematics problem — like the worked examples above — before attempting the textbook exercises.
All the chapter-wise sheets are collected in the Physics formulas hub if you need to jump between topics.
Why does the centre of mass of an exploding projectile continue along the same parabola?
The explosion forces are internal to the projectile, and internal forces cancel in pairs, so they cannot move the centre of mass. The only external force is gravity, which is the same before and after the explosion, so the centre of mass continues on the original parabolic path (NCERT, p. 99).
When can I write L = Iω for a rotating body?
For rotation about a fixed axis, the component along the axis is \( L_z = I\omega \). If the body is symmetric about the axis of rotation, the perpendicular components cancel and \( \mathbf{L} = I\omega\hat{\mathbf{k}} \) (NCERT, p. 121).
For a body not symmetric about the axis, \( \mathbf{L} \) is not along the axis, so only the axial component follows the simple formula.
How do I convert revolutions per minute to rad/s?
Multiply by \( 2\pi \) and divide by 60: \( \omega = \frac{2\pi\times\text{(rev/min)}}{60}\ \text{rad/s} \). NCERT’s Example 6.11 converts 1200 rpm to \( 40\pi\ \text{rad/s} \) this way (NCERT, p. 117). The full chapter text is available on the NCERT e-textbook portal (Physics Part I, Chapter 6, file keph106.pdf).
Reference: NCERT Class 11 Physics textbook, chapter Systems of Particles and Rotational Motion.
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- Systems of Particles and Rotational Motion Notes
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