Rotational dynamics about a fixed axis

Push on a heavy door near its hinge and it barely budges. Push the same force at the outer edge, far from the hinge, and it swings open easily. The force is identical — what changed is where you applied it. That single everyday fact is the seed of an entire branch of mechanics: rotational dynamics. Newton's second law, $F = ma$ , tells us how a force changes an object's straight-line motion, but it says nothing about spinning, and spinning is how wheels turn, fans blow, planets rotate and figure skaters pull in their arms to whirl faster. To describe turning we need a new quantity — torque — the rotational twin of force, and a new law that plays the role $F=ma$ plays for straight-line motion. This article builds that law from the ground up, using only the moments of inertia CBSE hands you ready-made (no integration required). We define torque as $\vec\tau = \vec r \times \vec F$ , show why only the perpendicular component of a force turns a body, and arrive at the rotational equation of motion $\tau_{net} = I\alpha$ — the direct rotational analogue of Newton's second law. Along the way we meet rotational kinetic energy $\tfrac12 I\omega^2$ , the two independent conditions a rigid body must satisfy to sit perfectly still (zero net force and zero net torque), and the classic trap of the "couple" — two forces that cancel in translation but not in rotation. By the end you will be able to compute the torque on a spinning wheel, find the force needed to balance a loaded rod, and read off angular acceleration the same way you read off linear acceleration from $F=ma$ .

Key takeaways

Torque is the rotational analogue of force. $\vec\tau = \vec r \times \vec F$ , with magnitude $\tau = rF\sin\theta$ , where $\theta$ is the angle between the position vector $\vec r$ (from the axis to the point of application) and the force $\vec F$ .
Only the perpendicular component of the force turns the body. A force pushing straight along $\vec r$ (through the axis) produces zero torque no matter how large it is.
The rotational equation of motion is $\tau_{net} = I\alpha$ — the exact rotational counterpart of $F_{net} = ma$ , with moment of inertia $I$ standing in for mass and angular acceleration $\alpha$ standing in for linear acceleration.
A spinning rigid body stores kinetic energy $K_{rot} = \tfrac12 I\omega^2$ ; a rolling body carries both this and translational KE, $K = \tfrac12 Mv_{cm}^2 + \tfrac12 I_{cm}\omega^2$ .
Equilibrium of a rigid body needs two separate conditions, not one: $\sum\vec F = 0$ (translational) and $\sum\vec\tau = 0$ (rotational) about any chosen point.
A couple is the classic trap — two equal, opposite, non-collinear forces give zero net force but a non-zero net torque, so a body can accelerate rotationally while its centre of mass stays put.
Key idea: torque is the moment of a force, $\tau = rF\sin\theta$ , and the rotational equation of motion $\tau_{net} = I\alpha$ governs how a rigid body's spin changes, exactly the way $F=ma$ governs straight-line motion.

The core insight

Torque $\vec\tau = \vec r \times \vec F$ measures a force's turning effect about an axis, and the net torque on a rigid body produces angular acceleration through $\tau_{net} = I\alpha$ , the rotational counterpart of $F=ma$ , with equilibrium of the body requiring both $\sum\vec F=0$ and $\sum\vec\tau=0$ .

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 Physics unit System of Particles and Rotational Motion, immediately after rotational kinematics (angular displacement, velocity and acceleration) and just before angular momentum. The prescribed scope is exactly torque $\vec\tau = \vec r\times\vec F$ , the equation of motion $\tau = I\alpha$ , rotational kinetic energy $\tfrac12 I\omega^2$ , and the two conditions for rigid-body equilibrium — using the moments of inertia of standard bodies (rod, disc, ring, sphere) as given formulas, never derived from integration, which is explicitly out of scope at this level. This article stays entirely within that boundary and connects backward to "Laws of Motion" (the linear $F=ma$ ) and forward to "Work, Energy and Power" (rotational kinetic energy) and to angular momentum, the next node in this chapter.

The physical picture

Consider a rigid body free to rotate about a fixed axis, and a force $\vec F$ applied at a point whose position relative to the axis is $\vec r$ . Experience with the door tells us two things matter: how large the force is, and how far from the hinge (and at what angle) it is applied. Torque packages both into a single vector quantity.

\vec\tau = \vec r \times \vec F, \qquad \tau = rF\sin\theta

where: $\vec r$ is the position vector from the axis (or chosen pivot) to the point where the force acts, $\vec F$ is the applied force, and $\theta$ is the angle between $\vec r$ and $\vec F$ when their tails are placed together. The physical reading of $\tau = rF\sin\theta$ is that only the component of $\vec F$ perpendicular to $\vec r$ actually turns the body — write $F\sin\theta$ as $F_\perp$ and torque is simply $rF_\perp$ , exactly analogous to how only the component of a force along the direction of motion does work. Torque is maximum ( $\tau = rF$ ) when the force is applied at right angles to $\vec r$ , and it is exactly zero when the force is directed straight along $\vec r$ , either pushing toward or pulling away from the axis — that force has no leverage at all. The direction of $\vec\tau$ , given by the right-hand rule, is along the rotation axis; in a fixed-axis problem we usually just track the sign (say, counter-clockwise positive) rather than the full vector.

Figure

A hinged door viewed from above, showing a force applied near the hinge producing a short lever arm and small torque, and the same size force applied at the outer edge producing a long lever arm and large torque, with the perpendicular distance from the hinge to the line of the force marked in each case

Once torque is defined, we can ask how it changes a body's spin. Just as a net force changes an object's velocity, a net torque about the fixed axis changes a body's angular velocity — and the constant of proportionality is the moment of inertia, the rotational analogue of mass.

\tau_{net} = I\alpha

where: $\tau_{net}$ is the sum of all external torques about the fixed axis, $I$ is the body's moment of inertia about that same axis ( $I=\sum m_ir_i^2$ for a system of particles, with ready-made formulas for standard shapes such as a rod, $\tfrac1{12}ML^2$ about its centre, a disc, $\tfrac12MR^2$ , a ring, $MR^2$ , and a solid sphere, $\tfrac25MR^2$ ), and $\alpha$ is the resulting angular acceleration. The physical reading mirrors $F=ma$ exactly: a larger torque produces a larger angular acceleration, and for the same torque, a body with a larger moment of inertia (mass spread farther from the axis) speeds up its spin more slowly — this is why a solid disc and a ring of the same mass and radius, given the same torque, accelerate differently, the ring (all its mass at the rim) lagging the disc (mass spread inward). A rotating body also stores energy in its spin, exactly as a moving body stores energy in its motion:

K_{rot} = \tfrac12 I\omega^2

where: $\omega$ is the angular velocity about the fixed axis. A body that both moves and spins — like a ball rolling down a slope — carries both kinds of energy at once, $K = \tfrac12 Mv_{cm}^2 + \tfrac12 I_{cm}\omega^2$ , the translational part tracking the centre of mass and the rotational part tracking the spin about that centre. Finally, a rigid body at rest, or moving without any change in its state, must satisfy two independent conditions: translational equilibrium, $\sum\vec F = 0$ , and rotational equilibrium, $\sum\vec\tau = 0$ about any point. These are genuinely separate requirements — a body can satisfy one without the other. The clearest example is a couple: two forces equal in magnitude, opposite in direction, acting along different (parallel) lines. Their vector sum is zero, so there is no net force and the centre of mass does not accelerate, yet each force has a non-zero lever arm about any point between them, so the torques add rather than cancel, and the body experiences a net turning effect. Spinning a steering wheel with two hands, one pushing up and the other pulling down, is exactly this.

Linear quantity	Rotational analogue
Force $F$	Torque $\tau$
Mass $m$	Moment of inertia $I$
$F_{net} = ma$	$\tau_{net} = I\alpha$
$K = \tfrac12 mv^2$	$K_{rot} = \tfrac12 I\omega^2$
Momentum $p = mv$	Angular momentum $L = I\omega$

Derivation AFrom torque to angular acceleration for a rigid body

The goal is to show why the single-particle statement "torque equals rate of change of angular momentum" collapses, for a rigid body rotating about a fixed axis with constant moment of inertia, into the compact and exam-ready form $\tau_{net} = I\alpha$ — the rotational twin of $F=ma$ — starting only from Newton's second law applied to one particle of the body and then summing over the whole body.

Step 1. Isolate one particle of the rigid body, of mass $m_i$ , sitting at perpendicular distance $r_i$ from the fixed axis. Because the body is rigid, every particle moves in a circle of its own radius $r_i$ but shares the same angular velocity $\omega$ and angular acceleration $\alpha$ as the whole body — that shared $\alpha$ is what makes a single equation for the whole body possible. The particle's tangential (speeding-up) acceleration is $a_{t,i} = r_i\alpha$ , and by Newton's second law the tangential force needed to produce it is $F_{t,i} = m_i r_i\alpha$ .

Step 2. Convert that tangential force into a torque about the axis. Since the tangential force acts perpendicular to $r_i$ , its torque about the axis is simply $\tau_i = r_i F_{t,i} = r_i(m_i r_i\alpha) = m_i r_i^2\alpha$ . This is the torque that internal and external forces together must supply to keep particle $i$ moving on its circle with angular acceleration $\alpha$ .

Step 3. Sum over every particle in the body. The internal forces between particles come in equal-and-opposite pairs along the line joining them (Newton's third law), so their torques about the axis cancel in pairs when added over the whole body — only the external torques survive the sum. Adding Step 2 over all particles: $\tau_{net} = \sum_i \tau_i = \sum_i m_i r_i^2\,\alpha = \left(\sum_i m_i r_i^2\right)\alpha$ .

Step 4. Recognise the bracketed sum as the moment of inertia of the body about the fixed axis, $I = \sum_i m_i r_i^2$ , by its very definition. Because every particle shares the same $\alpha$ , it comes outside the sum untouched, leaving $\tau_{net} = I\alpha$ — a single scalar equation, valid for the whole rigid body, that plays exactly the role $F_{net}=ma$ plays for a point mass, with $I$ replacing $m$ as the measure of "reluctance to change motion," now reluctance to change spin.

Worked examples

Board level - torque, angular acceleration and final angular velocity of a wheel

A wheel of moment of inertia $I = 0.6\ \text{kg}\,\cdot\,\text{m}^2$ is initially at rest. A constant tangential force of $12\ \text{N}$ is applied at the rim, at a perpendicular distance of $0.3\ \text{m}$ from the axis. Find (a) the torque, (b) the angular acceleration, and (c) the angular velocity after $4\ \text{s}$ .

Step 1. The force is applied perpendicular to the radius (tangential), so $\theta = 90^\circ$ and $\sin\theta = 1$ ; the torque is $\tau = rF = (0.3)(12) = 3.6\ \text{N}\,\cdot\,\text{m}$ .

Step 2. Apply the rotational equation of motion $\tau = I\alpha$ to get the angular acceleration: $\alpha = \dfrac{\tau}{I} = \dfrac{3.6}{0.6} = 6\ \text{rad/s}^2$ .

Step 3. Use the rotational kinematics equation $\omega = \omega_0 + \alpha t$ , with the wheel starting from rest ( $\omega_0 = 0$ ): $\omega = 0 + (6)(4) = 24\ \text{rad/s}$ . Answer: torque $= 3.6\ \text{N}\,\cdot\,\text{m}$ , angular acceleration $= 6\ \text{rad/s}^2$ , and the wheel is spinning at $24\ \text{rad/s}$ after $4\ \text{s}$ .

Competitive level - balancing a loaded rod and finding its rotational kinetic energy at release

A uniform rod of length $2\ \text{m}$ is pivoted at one end and held horizontal; the rod's own mass is negligible. A weight of $5\ \text{kg}$ hangs from the free end. A person applies an upward force $F$ at a point $0.5\ \text{m}$ from the pivot to keep the rod in rotational equilibrium. (a) Find $F$ . (b) If the person's supporting force is suddenly removed while the weight is at the free end, and the rod (treated for this part as a thin rod of mass $3\ \text{kg}$ and the same length, pivoted at the same end, with $I = \tfrac13ML^2$ about that end) starts rotating freely under gravity, find its angular acceleration at the instant of release.

Step 1. Take torques about the pivot. The hanging weight acts at $r = 2\ \text{m}$ with force $mg = (5)(9.8) = 49\ \text{N}$ , perpendicular to the rod, giving $\tau_{weight} = (2)(49) = 98\ \text{N}\,\cdot\,\text{m}$ , say clockwise.

Step 2. For rotational equilibrium the supporting force's torque, acting counter-clockwise at $r = 0.5\ \text{m}$ , must equal this: $\tau_F = (0.5)F = 98$ , so $F = \dfrac{98}{0.5} = 196\ \text{N}$ . A force of $196\ \text{N}$ is needed to hold the rod horizontal.

Step 3. For part (b), once the support is removed the only torque about the pivot is from gravity on the rod itself (its weight acts effectively at its centre, $r = 1\ \text{m}$ , since the $5\ \text{kg}$ mass is now treated separately and is not part of this simplified free-rotation check): $\tau = Mg\left(\dfrac{L}{2}\right) = (3)(9.8)(1) = 29.4\ \text{N}\,\cdot\,\text{m}$ , and the rod's moment of inertia about the pivot end is $I = \dfrac13 ML^2 = \dfrac13(3)(2)^2 = 4\ \text{kg}\,\cdot\,\text{m}^2$ .

Step 4. Apply $\tau = I\alpha$ : $\alpha = \dfrac{29.4}{4} = 7.35\ \text{rad/s}^2$ . Answer: $F = 196\ \text{N}$ to hold the rod in equilibrium; once released (rod alone), the initial angular acceleration is $7.35\ \text{rad/s}^2$ . Timing note: about two and a half minutes. The equilibrium half (Steps 1-2) is a single torque-balance equation and should take under a minute; the release half (Steps 3-4) is a fresh $\tau = I\alpha$ calculation using the given moment of inertia formula for a rod pivoted at one end — do not try to derive $I=\tfrac13ML^2$ from integration, it is simply quoted at this level. Keep straight which mass and which lever arm belong to which part of the question; mixing the $5\ \text{kg}$ weight's lever arm with the rod's own moment of inertia is the single most common slip.

Common traps

Forgetting torque depends on the angle between $\vec r$ and $\vec F$ — only the perpendicular component of the force contributes; a force directed straight along $\vec r$ produces zero torque no matter how large.
Ignoring the sign convention for torques — torques that would rotate the body clockwise and counter-clockwise about the chosen point must be assigned opposite signs before summing to zero for equilibrium.
Assuming zero net force automatically means zero net torque, or vice versa — the two equilibrium conditions are independent; a couple is the standard counterexample, giving zero net force but a non-zero net torque.
Using $\tau = I\alpha$ with $\tau$ and $I$ computed about different axes — both quantities in the equation must refer to the exact same fixed axis, chosen once and used consistently.
Trying to derive a moment of inertia formula (rod, disc, ring, sphere) from scratch by integration — at this level these are given formulas to be quoted and applied, not derived.
Confusing rotational kinetic energy $\tfrac12I\omega^2$ with translational kinetic energy $\tfrac12Mv^2$ for a rolling body — a rolling object has both simultaneously and the two must be added, not chosen between.

Common trapA couple can spin a body with zero net force — do not assume force balance implies torque balance

The single biggest mark-losing habit in this topic is treating "the forces cancel" as proof that a body is in complete equilibrium. It is not. A couple — two equal, opposite, non-collinear forces, such as two hands turning a steering wheel in opposite directions — produces exactly zero net force, so the centre of mass does not accelerate at all, yet it produces a very real, non-zero net torque, because each force has a non-zero perpendicular distance from any chosen point and their torques add instead of cancelling. Complete rigid-body equilibrium genuinely requires checking both conditions separately: $\sum\vec F=0$ for translation and $\sum\vec\tau=0$ , computed about any convenient point, for rotation. Never infer one from the other.

You should be able to…

Compute torque as $\tau = rF\sin\theta$ for a given force and its point of application about a chosen axis.
Apply the rotational equation of motion $\tau_{net} = I\alpha$ to find angular acceleration for a rigid body, quoting the correct given moment-of-inertia formula for the shape involved.
State and correctly apply both independent conditions for rigid-body equilibrium: net force zero and net torque zero about any point, and recognise a couple as the case where the first holds without the second.
Compute rotational kinetic energy $\tfrac12I\omega^2$ and combine it with translational kinetic energy for a rolling or otherwise translating-and-spinning body.
Use the linear-rotational analogy table to translate a familiar $F=ma$ -style problem into its rotational version, and vice versa.
Set up a torque balance about a sensibly chosen pivot point to solve equilibrium problems involving hanging weights and supporting forces.

Ask the tutor: Want help choosing the best pivot point for a torque-balance problem, or working through a few more $\tau = I\alpha$ numericals with different standard shapes?

Why is torque a vector and not just a number like force magnitude times distance?

Torque genuinely has a direction, not just a size, because it tells you which way something turns, not just how hard. For fixed-axis problems, like almost everything in this chapter, you can get away with treating it as a signed number — positive for counter-clockwise, negative for clockwise, say — because the rotation axis itself never changes direction. The full vector definition $\vec\tau = \vec r\times\vec F$ becomes essential once a body can rotate about more than one axis at once, which is beyond the Class 11 syllabus; for now, just fix a sign convention at the start of a problem and stick to it consistently.

If I push a door at the very edge, why does that take less force to open than pushing near the hinge?

Because torque is $\tau = rF\sin\theta$ , and for the same push perpendicular to the door, a larger $r$ (distance from the hinge) gives a larger torque for the same force. To produce the same torque needed to swing the door, a smaller force suffices when $r$ is large. That is precisely why door handles sit far from the hinge and not right next to it — it is a deliberate application of the lever-arm idea, letting you supply less force for the same turning effect.

Do I need to memorise the derivation of tau = I alpha for the board exam, or just the formula?

For CBSE Class 11, you are expected to use $\tau_{net} = I\alpha$ confidently in numericals — computing torque, moment of inertia from the given standard formulas, and angular acceleration — rather than reproduce a full derivation from Newton's third law and internal-force cancellation from memory. The derivation shown here is worth understanding once, because it builds real intuition for why $I$ plays the role of "rotational mass," but if your board paper asks a numerical, go straight to the formula and the given moment-of-inertia value for the shape in question.

Why does a ring lag behind a disc of the same mass and radius when the same torque is applied to both?

Because their moments of inertia differ even though their mass and radius are identical: a ring has all its mass at the rim, $I_{ring} = MR^2$ , while a disc has its mass spread from the centre outward, $I_{disc} = \tfrac12MR^2$ — half as much. Since $\alpha = \tau/I$ for the same torque, the disc, with the smaller $I$ , gets a larger angular acceleration and spins up faster. This is the direct rotational parallel to how, for the same force, a lighter object accelerates faster than a heavier one under $F=ma$ — here it is not how much mass there is that matters most, but how far that mass sits from the axis.