Switch off a ceiling fan on a hot afternoon in Chennai. The blades do not stop at once. They keep turning for most of a minute, slowing gently all the while. Nothing pushes them forward any more, and nothing grabs them to a halt. Some small, steady effect in the bearings and the air removes their spin, bit by bit. A heavier fan, or a faster one, takes longer to stop.
Adding up the particles
A rigid body turning about a fixed axis is a crowd of particles, each moving on its own circle about that axis. From lesson 1, a particle of mass \(m\) at distance \(r\) from the axis has angular momentum \(mr^2\omega\) about the axis. Every particle of a rigid body has the same \(\omega\), and every contribution points the same way along the axis. So the contributions add as plain numbers.
\[ L = m_1 r_1^2\omega + m_2 r_2^2\omega + m_3 r_3^2\omega + \dots \]
Why. each particle moves on a circle about the axis, and from lesson 1 it has \(L = mr^2\omega\) about the axis.
\[ L = \left(\sum m_i r_i^2\right)\omega \]
Why. \(\omega\) is the same for every particle of a rigid body, so it comes out of the sum.
\[ L = I\omega \]
Read it. \(\sum m_i r_i^2\) is the moment of inertia \(I\) about the axis, from the chapter on Rigid Bodies and Moment of Inertia.
The result is the rotational twin of \(p = mv\). The moment of inertia \(I\) takes the place of mass, and the angular velocity takes the place of velocity. For a continuous body the sum becomes an integral, \(L = \omega\int r^2\,dm = I\omega\). The moment of inertia found in the chapter on Rigid Bodies and Moment of Inertia is exactly the quantity needed here.
A rigid body turning with angular velocity \(\omega\) about a fixed axis has angular momentum \(L = I\omega\) about that axis. Here \(I\) is its moment of inertia about the same axis.
A uniform grinding disc of mass \(4\,\text{kg}\) and radius \(0.3\,\text{m}\) spins at \(10\,\text{rad/s}\) about its central axis. Find its angular momentum.
For a uniform disc about its central axis, \(I = \tfrac12 MR^2 = \tfrac12 \times 4 \times 0.3^2 = 0.18\,\text{kg m}^2\).
\(L = I\omega = 0.18 \times 10 = 1.8\,\text{kg m}^2\text{/s}\), along the axis. By the right-hand rule it points out of the face from which the disc is seen turning anticlockwise.
The vector form \(\vec L = I\vec\omega\) is safe when the axis is an axis of symmetry of the body. Every wheel, disc and rod in this chapter qualifies. For a lopsided body turning about a fixed axis, \(\vec L\) need not lie along the axis at all. Its component along the axis is still \(I\omega\), and that is the only component a fixed-axis problem needs. The bearings supply the torque that keeps the rest of \(\vec L\) turning with the body, which is why an unbalanced wheel shakes its axle.
A thin ring and a uniform disc have the same mass \(M\) and the same radius \(R\). Both spin at the same \(\omega\) about their central axes. How do their angular momenta compare?
A uniform rod of mass \(3\,\text{kg}\) and length \(1\,\text{m}\) turns at \(4\,\text{rad/s}\) about an axis through one end, perpendicular to the rod. Find its angular momentum about that axis, in kg m²/s.
What changes angular momentum
Newton's second law in its general form says \(\vec F = d\vec p/dt\): a force changes momentum. What changes angular momentum? The answer is the torque of the force about the same point, \(\vec\tau = \vec r\times\vec F\), the moment of the force. The chapter on Torque and Equilibrium studies it in detail; here one line of it is enough. Differentiate \(\vec L = \vec r\times\vec p\) with the product rule, keeping the order of each cross product.
\[ \dfrac{d\vec L}{dt} = \dfrac{d\vec r}{dt}\times\vec p + \vec r\times\dfrac{d\vec p}{dt} \]
Why. the product rule holds for a cross product, provided the order of the factors is kept.
\[ \dfrac{d\vec r}{dt}\times\vec p = \vec v\times m\vec v = 0 \]
Why. \(d\vec r/dt = \vec v\) and \(\vec p = m\vec v\) point the same way, and the cross product of parallel vectors is zero.
\[ \vec r\times\dfrac{d\vec p}{dt} = \vec r\times\vec F = \vec\tau \]
Why. \(d\vec p/dt = \vec F\), the net force, by Newton's second law.
\[ \dfrac{d\vec L}{dt} = \vec\tau \]
Read it. only the second term survives, and it is the torque of the net force about O.
The net torque on a particle about a fixed point O equals the rate of change of its angular momentum about O: \(\vec\tau = d\vec L/dt\).
This is the rotational twin of \(\vec F = d\vec p/dt\), and it explains lesson 1 at once. A particle moving in a straight line at constant velocity feels no force, so it feels no torque about any point. Its angular momentum about every point must therefore stay constant, which is exactly what the figure of lesson 1 showed.
A system, and a body on a fixed axis
Now add the torque law over every particle of a system. Each particle feels external forces and internal forces from the other particles. By Newton's third law the internal forces come in equal and opposite pairs. The two forces of a pair also act along the same line, the line joining the two particles. So they have the same moment arm about any point, and their torques cancel. Only the torques of the external forces survive.
For a rigid body on a fixed axis, \(L = I\omega\) with \(I\) constant. Then \(dL/dt = I\,d\omega/dt = I\alpha\), and the torque law becomes \(\tau = I\alpha\). That is the law of the chapter on Dynamics of Rotation. The form \(\tau = dL/dt\) is the more general one. When a spinning body changes shape, \(I\) changes, and \(\tau = I\alpha\) fails. The law \(\tau = dL/dt\) still holds, and lesson 4 is built on it.
Which of these can change the total angular momentum of a system about a fixed point O?
A spinning body changes its shape, so its moment of inertia changes with time. Which relation still holds throughout?
A projectile checks the law
The torque law is not only about wheels. Take a ball thrown from a point O on level ground, with velocity components \(u_x\) across and \(u_y\) up. Take O as the fixed point. The only force is the weight \(mg\), acting straight down through the ball. Its line of action is vertical, so its moment arm about O is the horizontal distance \(x\) of the ball from O.
A ball of mass \(m\) is thrown from O with velocity components \(u_x\) and \(u_y\). Take x across, y up and z out of the page. Show that \(\tau_z = dL_z/dt\) about O at every instant, and find both at \(t = 1\,\text{s}\) for \(m = 0.2\,\text{kg}\), \(u_x = 6\,\text{m/s}\), \(u_y = 8\,\text{m/s}\), \(g = 10\,\text{m/s}^2\).
Position: \(x = u_xt\), \(y = u_yt - \tfrac12 gt^2\). Velocity: \(v_x = u_x\), \(v_y = u_y - gt\). Force: \(F_x = 0\), \(F_y = -mg\).
Torque: \(\tau_z = xF_y - yF_x = -mgx\). Angular momentum: \(L_z = m(xv_y - yv_x)\).
\(L_z = m\left[u_xt(u_y - gt) - u_x(u_yt - \tfrac12 gt^2)\right] = -\tfrac12 mgu_xt^2\). Its rate is \(dL_z/dt = -mgu_xt = -mgx\), which equals \(\tau_z\).
At \(t = 1\,\text{s}\): \(x = 6\,\text{m}\), so \(\tau_z = -0.2 \times 10 \times 6 = -12\,\text{N m}\), and \(L_z = -\tfrac12 \times 0.2 \times 10 \times 6 \times 1 = -6\,\text{kg m}^2\text{/s}\). Both are clockwise, and \(L\) grows at exactly the rate the torque demands.
A \(0.1\,\text{kg}\) ball is thrown from O with \(u_x = 10\,\text{m/s}\) and \(u_y = 10\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\). Find the magnitude of its angular momentum about O at \(t = 2\,\text{s}\), in kg m²/s.
A projectile moves under gravity alone. Is its angular momentum about the launch point O conserved during the flight?
Newton's Principia of 1687 gave laws for the motion of particles, not of spinning bodies. For almost a century mechanics treated rotation by special tricks, one problem at a time. In 1775 the Swiss mathematician Leonhard Euler stated the torque law as a principle of its own. The rate of change of angular momentum equals the applied torque, just as the rate of change of momentum equals the applied force. Engineers still call the pair Euler's laws of motion.
Return to the ceiling fan. Switched off, it feels no driving torque, only a small friction torque from the bearings and the air. That torque removes its angular momentum at a steady rate. A fan with \(I = 0.3\,\text{kg m}^2\) at \(300\) revolutions per minute has \(\omega = 31.4\,\text{rad/s}\) and \(L = 9.4\,\text{kg m}^2\text{/s}\). If it stops in \(60\,\text{s}\), the average friction torque is only \(9.4/60 \approx 0.16\,\text{N m}\). A larger \(I\) or a larger \(\omega\) means more \(L\) to remove, so the fan runs longer.
- Using \(\tau = I\alpha\) for a body that changes its shape.Use \(\tau = dL/dt\). The form \(I\alpha\) needs a constant \(I\).
- Taking the torque about one point and the angular momentum about another.Both sides of \(\tau = dL/dt\) must be about the same fixed point or the same fixed axis.
- Counting internal forces in the torque law.Internal forces act in pairs along one line, so their torques cancel. Count only external torques.