A watchman sits at the gate of a housing society in Pune, a few metres back from a straight road. A car goes past at a steady speed. When the car is far away, the watchman barely moves his head to follow it. As the car comes level with the gate, his head swings round quickly. Then it slows again as the car moves away down the road. The car's speed never changed, yet its turning about the watchman changed all the time. Something about the car's motion, measured from the watchman, stays exactly the same from start to end.
From momentum to its moment
The momentum \(\vec p = m\vec v\) of a particle describes its motion along its path. It says nothing about how the particle turns about some chosen point. For that we need a new quantity, built the way torque is built from force. The torque of a force about a point is \(\vec\tau = \vec r\times\vec F\), as the chapter on Torque and Equilibrium develops. The same product with momentum in place of force is the angular momentum.
Choose a fixed point O. Let \(\vec r\) be the position vector of the particle from O, and \(\vec p\) its momentum. The angular momentum of the particle about O is the cross product \(\vec L = \vec r\times\vec p\). Its magnitude is \(L = rp\sin\theta = mvr\sin\theta\), where \(\theta\) is the angle between \(\vec r\) and \(\vec p\) placed tail to tail.
The angular momentum of a particle about a fixed point O is \(\vec L = \vec r\times\vec p\). Here \(\vec r\) is the position vector of the particle from O and \(\vec p = m\vec v\) is its momentum. Its SI unit is kg m²/s.
Direction and the choice of point
A cross product is perpendicular to both of its vectors. So \(\vec L\) is perpendicular to the plane that contains \(\vec r\) and \(\vec p\). Its sense follows the right-hand rule. Curl the fingers of the right hand from \(\vec r\) towards \(\vec p\), and the thumb points along \(\vec L\). For motion in the page, \(\vec L\) points out of the page when the particle goes round O anticlockwise. It points into the page when the particle goes round clockwise.
Angular momentum always belongs to a chosen point. Move the point and \(\vec r\) changes, so \(\vec L\) changes too. One particle has a different angular momentum about every point. A value of \(L\) that names no point is as incomplete as a moment of inertia that names no axis.
Angular momentum does not need rotation. A particle has angular momentum about O whenever its line of motion misses O, even if it moves in a perfectly straight line. It is zero only when the momentum points straight towards O or straight away from it, or when the particle is at O.
A particle moves at constant velocity along a straight line. Point A lies on that line, and point B lies \(2\,\text{m}\) from it. What can you say about its angular momentum about A and about B?
Two groupings of one product
The product \(rp\sin\theta\) can be grouped in two ways, and each grouping is a method. First, group \(\sin\theta\) with \(p\). The part of the momentum across \(\vec r\) is \(p_\perp = p\sin\theta\), and the part along \(\vec r\) adds nothing. Then \(L = rp_\perp\). Second, group \(\sin\theta\) with \(r\). The length \(d = r\sin\theta\) is the perpendicular distance from O to the line of motion. It is called the moment arm, and \(L = pd\).
The second method is usually the quicker one. Draw the line of motion, drop a perpendicular onto it from O, and multiply its length by the momentum. The same two methods give torque in the chapter on Torque and Equilibrium, with force in place of momentum. A third method uses components. For motion in the xy plane with O at the origin, \(L_z = xp_y - yp_x\), with a positive value meaning out of the page.
A \(0.5\,\text{kg}\) ball is at the point \((4\,\text{m}, 3\,\text{m})\) and moves with velocity \((2\,\text{m/s}, 4\,\text{m/s})\). Find its angular momentum about the origin O.
By components: \(p_x = 0.5 \times 2 = 1\,\text{kg m/s}\) and \(p_y = 0.5 \times 4 = 2\,\text{kg m/s}\). Then \(L_z = xp_y - yp_x = 4 \times 2 - 3 \times 1 = 5\,\text{kg m}^2\text{/s}\), out of the page.
By \(rp\sin\theta\): \(r = 5\,\text{m}\) and \(p = 0.5\sqrt{20} = 2.24\,\text{kg m/s}\). The angle of \(\vec r\) is \(36.9°\) and that of \(\vec v\) is \(63.4°\), so \(\theta = 26.6°\) and \(\sin\theta = 0.447\).
Then \(L = 5 \times 2.24 \times 0.447 = 5.0\,\text{kg m}^2\text{/s}\), the same as before.
The positive sign means the ball goes round O anticlockwise at this instant, so \(\vec L\) points out of the page.
A particle has momentum \(8\,\text{kg m/s}\) at a distance of \(3\,\text{m}\) from a point O. The angle between \(\vec r\) and \(\vec p\) is \(30°\). Find its angular momentum about O, in kg m²/s.
A \(2\,\text{kg}\) particle is at \((1\,\text{m}, 2\,\text{m})\) and has velocity \((3\,\text{m/s}, -1\,\text{m/s})\). Find the magnitude of its angular momentum about the origin, in kg m²/s.
A particle moves at a steady speed along a straight line that passes \(3\,\text{m}\) from a point O. As it approaches O, passes it and moves away, what happens to its angular momentum about O?
The particle in the figure above has a mass of \(0.5\,\text{kg}\) and a speed of \(4\,\text{m/s}\). Far to the left, \(r\) is about \(7\,\text{m}\) and \(\theta\) is close to \(180°\). When the particle is level with O, \(r = 3\,\text{m}\) and \(\theta = 90°\). At every instant \(r\sin\theta = 3\,\text{m}\), because it is the moment arm, the fixed distance from O to the line. So \(L = mvd = 0.5 \times 4 \times 3 = 6\,\text{kg m}^2\text{/s}\), into the page, all the way.
Now choose the chip that puts O on the line itself. The moment arm is zero, and so \(L = 0\) at every instant. The same motion has a constant angular momentum of \(6\,\text{kg m}^2\text{/s}\) about one point and zero about another. The constancy is no accident. No force acts on the particle, so no torque acts about any point, and lesson 2 shows that without a torque \(L\) cannot change.
A particle moving at constant velocity \(\vec v\) has a constant angular momentum \(L = mvd\) about any fixed point. Here \(d\) is the perpendicular distance from the point to its line of motion.
A cricket ball of mass \(0.16\,\text{kg}\) rolls across the outfield at a steady \(5\,\text{m/s}\), along a line that passes \(3\,\text{m}\) from a fielder's feet. Find its angular momentum about the fielder's feet when it is level with him, and again when it is \(12\,\text{m}\) further on.
The moment arm is the distance from the feet to the line of motion, \(d = 3\,\text{m}\), at every instant.
Level with the fielder, \(L = mvd = 0.16 \times 5 \times 3 = 2.4\,\text{kg m}^2\text{/s}\).
Twelve metres on, \(r\) has grown to \(\sqrt{12^2 + 3^2} = 12.4\,\text{m}\), but \(r\sin\theta\) is still \(3\,\text{m}\). So \(L\) is still \(2.4\,\text{kg m}^2\text{/s}\). Friction and air drag are ignored here.
In which of these is the angular momentum of the particle about the point O zero?
The particle on a circle
For a particle moving on a circle about its centre O, the velocity is always along the tangent. The tangent is perpendicular to the radius, so \(\theta = 90°\) and \(\sin\theta = 1\) at every point. Then \(L = mvr\). With \(v = \omega r\) from the chapter on Kinematics of Circular Motion, \(L = mr^2\omega\). The vector \(\vec L\) lies along the axis of the circle, in the same direction as \(\vec\omega\).
The form \(mr^2\omega\) should look familiar. Each particle of a spinning body contributes exactly this. The sum of \(mr^2\) over the body is its moment of inertia, from the chapter on Rigid Bodies and Moment of Inertia. The next lesson adds these contributions for a whole rigid body.
A \(0.2\,\text{kg}\) stone on a string is whirled on a horizontal circle of radius \(0.5\,\text{m}\) at \(3\,\text{rad/s}\). Find its angular momentum about the centre of the circle, in kg m²/s.
A \(30\,\text{kg}\) child sits at the rim of a merry-go-round at a village fair, \(2\,\text{m}\) from the centre. The ride makes one turn in \(4\,\text{s}\). Estimate the child's angular momentum about the axle, in kg m²/s.
Now return to the watchman at the gate. Take the car as a particle and the watchman as the point O. The car's angular momentum about him is \(mvd\), where \(d\) is his distance from the road, and it never changes. Write it also as \(mr^2\omega\), where \(\omega\) is the rate at which his head turns. Then \(\omega = vd/r^2\). Take a car at \(10\,\text{m/s}\) and a gate \(5\,\text{m}\) from the road. His head turns at \(2\,\text{rad/s}\) when the car is level, and at only \(0.02\,\text{rad/s}\) when the car is \(50\,\text{m}\) away.
- Saying that a particle moving in a straight line has no angular momentum.Use \(L = pd\). It is zero only about points on the line of motion.
- Quoting an angular momentum without naming the point.Always say "about O". The same particle has a different \(L\) about every point.
- Using the distance \(r\) from O in place of the moment arm \(d\).The moment arm is measured at right angles from O to the line of motion, \(d = r\sin\theta\).