Rigid body mechanics

For rotation about a fixed axis, constant-angular-acceleration equations mirror the familiar linear equations. The mapping is $s\leftrightarrow\theta$, $u\leftrightarrow\omega_0$, $v\leftrightarrow\omega$, and $a\leftrightarrow\alpha$.

The equations apply only while angular acceleration is constant. Before substituting, state a positive rotational direction and attach signs to angular velocities and angular acceleration. A rotating object can slow down while its angular velocity is positive if its angular acceleration is negative.

Angular-displacement graphs use the same gradient ideas as linear-motion graphs: the gradient of a $\theta$-against-$t$ graph is $\omega$, while the gradient of an $\omega$-against-$t$ graph is $\alpha$. The signed area under an $\omega$-against-$t$ graph is angular displacement.

A force can change translational motion and also produce a turning effect. Torque about a chosen point or axis measures this turning effect. It depends on the force, the distance from the axis, and the angle between the position vector and the force.

For forces in a plane perpendicular to the chosen axis, the torque magnitude about that axis is $\tau=rF\sin\phi=Fd_\perp$. Here $r$ runs from the pivot to the force point in that plane, and $d_\perp$ is the perpendicular distance from the pivot to the force's line of action. Assign the rotational sign separately. A force whose line of action passes through the axis gives zero torque about that axis. In three dimensions, torque about an axis is the component of the torque vector along that axis, not generally the vector's full magnitude.

Torque direction follows the right-hand rule. In a two-dimensional problem, assign clockwise and counterclockwise torques opposite signs. A couple is a pair of equal, opposite, parallel forces with different lines of action: its resultant force is zero but its torque is not.

Qualitative schematics. Left: the force is perpendicular to the beam, so its lever arm is r. Centre: equal loads act at equal distances from the pivot of an ideal weightless beam; the upward reaction balances both loads. Right: rolling without slipping on stationary ground; speeds are measured relative to the ground.Axis-first checklist
Torque: Draw the force's complete line of action, then the shortest perpendicular from the pivot to that line. Its length is $d_\perp$, giving $|\tau|=Fd_\perp=rF\sin\phi$ for $0\le\phi\le\pi$. A downward force applied to the right of a pivot produces clockwise torque; a downward force to its left produces counterclockwise torque.
Equilibrium: Two downward loads on opposite sides balance rotationally when $F_1d_1=F_2d_2$, provided no other moment acts. Full static equilibrium also requires $\sum\limits\vec F=0$; for an ideal weightless beam, the upward support force is $N=F_1+F_2$.
Rolling: A wheel rolling right without slipping on a stationary surface rotates clockwise. Its centre speed is $v_{\mathrm{CM}}=|\omega|R$; the instantaneous contact point has zero velocity relative to that surface.

In the planar models used here, complete mechanical equilibrium requires both zero resultant external force and zero resultant external torque: $\mathbf F_{\mathrm{net}}=\mathbf0$ and $\tau_{\mathrm{net}}=0$. Force balance prevents centre-of-mass acceleration; torque balance about the axis perpendicular to the plane prevents angular acceleration for constant $I$. When the resultant force is zero, changing the origin does not change the resultant torque.

In a uniform gravitational field, the resultant weight can be treated as acting through the centre of mass, which then coincides with the centre of gravity. Choosing a pivot through an unknown support force often eliminates that force's torque and simplifies the equation.

Under gravity alone, a supported body's limiting balance occurs when the vertical through its centre of mass reaches an edge of its base; beyond that edge, its weight produces an overturning torque. A wider base and lower centre of mass generally improve this gravitational stability. If additional forces act, find the tipping threshold by balancing all torques about the candidate base edge: an applied push can cause tipping while the centre-of-mass projection is still inside the base.

The moment of inertia $I$ measures how strongly a body's mass distribution resists angular acceleration about a specified axis. It is the rotational analogue of mass, but it is not determined by total mass alone: moving mass farther from the axis increases $I$ in proportion to $r^2$.

For point masses, add $m_ir_i^2$. For a continuous body, the same idea becomes $I=\int r^2\,dm$. Always state the axis because the same object has different moments of inertia about different axes.

IB guidance states that the moment-of-inertia expression for a specific mass distribution will be supplied when needed. Use it with the stated shape and axis; the values below are reference examples, not an additional memorization list. Calculating $I$ for point masses remains part of the core. The continuous-body integral and parallel-axis theorem explain useful relationships between inertia results. The parallel-axis theorem, $I=I_{\mathrm{CM}}+Md^2$, shifts a known value about a centre-of-mass axis to a parallel axis a distance $d$ away.

For a rigid body rotating about a fixed axis, the rotational form of Newton's second law is $\tau_{\mathrm{net}}=I\alpha$. The left side is the resultant external torque, not one selected applied torque. A larger $I$ produces a smaller angular acceleration for the same net torque.

Solve rotational-dynamics problems as you solve linear ones: isolate the body, draw all external forces, choose the axis and positive sense, calculate each signed torque, and only then connect the resultant torque to angular acceleration.

The resultant external torque acting over time provides the net angular impulse. About the same fixed axis, $\int\tau_{\mathrm{ext,net}}\,dt=\Delta L$, where $L$ is the signed angular-momentum component about that axis. For constant net torque, $\tau_{\mathrm{ext,net}}\Delta t=\Delta L$; for a varying net torque, use its time-average in this product. The signed area under a net external torque–time graph therefore equals the change in the system's angular momentum about the axis.

Angular momentum is defined relative to a chosen origin or axis. For a particle, the optional vector form is $\mathbf L=\mathbf r\times\mathbf p$, with $\mathbf r$ measured from that origin. For rotation about a fixed axis, $L=I\omega$ gives the signed component of angular momentum along that axis. About a principal axis, the angular-momentum and angular-velocity vectors are parallel, so the full angular-momentum magnitude is $I|\omega|$. In general three-dimensional rotation these vectors need not be parallel.

About a fixed origin in an inertial frame, external torque changes angular momentum: $\boldsymbol\tau_{\mathrm{ext,net}}=\frac{d\mathbf L}{dt}$. If the resultant external torque is zero, the total angular-momentum vector is conserved; a negligible torque gives approximate conservation. For the fixed-axis scalar model, zero net external torque about the axis conserves the total signed angular-momentum component about that axis. Internal interactions can transfer angular momentum between parts without changing the system total.

Conservation of angular momentum does not automatically imply conservation of kinetic energy. When rotating objects stick together, angular momentum can be conserved while mechanical energy is transferred to internal energy, sound, and deformation.

For rotation about a fixed axis, use the signed torque component about that axis. A constant torque does work $W=\tau\Delta\theta$, which is negative when the torque opposes the rotation. More generally, rotational work is the signed area under a torque–angle graph: $W=\int\tau\,d\theta$.

For a rigid body rotating about a fixed axis, $K_{\mathrm{rot}}=\tfrac12I\omega^2=\frac{L^2}{(2I)}$, where $L=I\omega$ is the signed angular-momentum component about that axis. With constant $I$, the net work of the external torques about the axis equals the change in rotational kinetic energy. This energy method can connect initial and final states without first finding angular acceleration or time.

Rotational power is the rate of doing rotational work. For rotation about a fixed axis, $P=\tau\omega$ uses signed torque and angular velocity about that axis; a resisting torque gives negative power. For a real machine, transfers into unintended pathways such as heating, sound, and deformation reduce useful mechanical-output efficiency below $100\%$.

An axisymmetric rigid wheel or cylinder rolling along a straight, stationary surface has translational kinetic energy of its centre of mass and rotational kinetic energy about that centre. For rolling without slipping, $v_{\mathrm{CM}}=R|\omega|$, where $v_{\mathrm{CM}}$ is speed. Choose matching positive linear and rotational senses so the signed tangential accelerations satisfy $a_{\mathrm{CM}}=R\alpha$. For free gravity-driven rolling down a fixed incline, the acceleration formula below also assumes no applied driving torque, negligible rolling resistance, and sufficient static friction to prevent slipping.

For rolling without slipping on a stationary surface, the material point at contact is instantaneously at rest relative to that surface. The point diametrically opposite contact—at the top of the wheel when the surface is horizontal—has speed $2v_{\mathrm{CM}}$ relative to the ground. These values come from vector addition of the centre-of-mass velocity and the velocity due to rotation about the centre.

Static friction can provide the torque needed to start or maintain rolling without dissipating mechanical energy at an ideal, non-slipping contact. Its direction is determined from the tendency to slip, not by memorizing that friction always points uphill or downhill.

Many HL problems couple a translating object to a rotating pulley, disk, or wheel. The string tension may be different from the hanging weight because some of the weight's effect accelerates the mass and some produces angular acceleration of the rotating body.

Use the connection constraint appropriate to the model. A taut string that does not slip on a fixed-axis pulley has tangential acceleration $a=\alpha R$ when positive string and pulley senses are matched. A wheel rolling without slipping along a straight, stationary surface has $v_{\mathrm{CM}}=R|\omega|$. Apply linear dynamics to translating parts, rotational dynamics to rotating parts, and keep the axis and sign convention consistent.

A strong IB solution begins with a system boundary and axis, a force diagram for each body, and a list of constraints. It then uses symbolic equations before numerical substitution, carries guard digits, includes units, and checks limiting cases such as a negligible pulley inertia.