At a busy railway station a passenger pulls a trolley bag along the platform and meets a raised edge at the end. The small wheels strike the edge with a jolt. Sometimes they hop up onto it; sometimes they stop dead and the bag tips. The edge pushes on the wheel with a large force for a moment, and nobody knows how large. Yet the motion just after the jolt can be found exactly.
Two parts of one angular momentum
The chapter on Angular Momentum defined \(\vec L = \vec r \times \vec p\) for a particle and \(L = I\omega\) for a body turning about a fixed axis. A rolling wheel does both things at once. Its centre moves, and it spins about its centre. Its angular momentum about a fixed point \(O\) must include both motions.
\[ \vec r_i = \vec r_{cm} + \vec r_i\,', \qquad \vec v_i = \vec v_{cm} + \vec v_i\,' \]
Why. write the position and velocity of each particle as those of the centre of mass plus a part measured from the centre of mass.
\[ \vec L_O = \sum m_i(\vec r_{cm} + \vec r_i\,')\times(\vec v_{cm} + \vec v_i\,') \]
Why. add \(\vec r\times m\vec v\) over all the particles of the body.
\[ \sum m_i\vec r_i\,' = 0, \qquad \sum m_i\vec v_i\,' = 0 \]
Why. seen from the centre of mass, the centre of mass is at the origin and at rest, so the two cross terms vanish.
\[ \vec L_O = \vec r_{cm}\times M\vec v_{cm} + \sum m_i\,\vec r_i\,'\times\vec v_i\,' \]
Why. the first term uses \(\sum m_i = M\); the last term is the angular momentum seen from the centre of mass.
\[ \vec L_O = \vec r_{cm}\times M\vec v_{cm} + I_{cm}\vec\omega \]
Read it. seen from the centre of mass the body only turns, so that last term is \(I_{cm}\vec\omega\).
The first term is the orbital angular momentum. It is what the body would have if all its mass sat at the centre of mass and moved with it. By the moment-arm rule of the chapter on Angular Momentum, its size is \(Mv_{cm}d\). The second term is the spin angular momentum, the spin measured about the centre of mass itself. The spin term does not depend on \(O\) at all. The orbital term does.
Name the point, then add with signs
The kinetic energy of a rolling body also splits into two parts, \(\tfrac12 Mv_{cm}^2\) and \(\tfrac12 I_{cm}\omega^2\). Those are numbers, and they always add. The two parts of angular momentum are different. Each points along the axis, either clockwise or anticlockwise in the plane of the figure. When their senses agree they add, and when they disagree they subtract. The total can even be smaller than either part.
Work through the chips of Fig 5.1. About a point \(O\) on the ground, the centre moves right along a line \(R\) above \(O\), so the orbital term is \(8 \times 3 \times 0.5 = 12\,\text{kg m}^2\text{/s}\), clockwise. The rolling disc also spins clockwise with \(I_{cm}\omega = \tfrac12 \times 8 \times 0.25 \times 6 = 6\). The total is \(18\,\text{kg m}^2\text{/s}\), clockwise. Reverse the spin and the total falls to \(12 - 6 = 6\).
About the centre itself, \(d = 0\) and only the spin remains: \(6\,\text{kg m}^2\text{/s}\). About a point \(T\) level with the top of the disc, the centre passes below \(T\), so the orbital term is \(12\) anticlockwise. With the spin of \(6\) clockwise, the total is \(6\,\text{kg m}^2\text{/s}\) anticlockwise. One body, one instant, and three different answers. An angular momentum without its point means nothing.
A body moves in a straight line and spins. About which fixed points does it have the same angular momentum?
A rolling body, counted about its contact
For a body rolling without slipping, take the point \(P\) on the ground just below the centre, where the body touches it at this instant. The orbital term is \(MvR\) and the spin term is \(I_{cm}\omega\), both clockwise for rolling to the right. With \(v = \omega R\) the total becomes \((I_{cm} + MR^2)\omega\). The bracket is the moment of inertia about the contact, \(I_P\), by the parallel-axis theorem.
This is the angular momentum version of a fact from the chapter on Rolling Motion. At each instant a rolling body turns purely about its contact. For the disc of Fig 5.1, \(I_P = 1 + 8 \times 0.25 = 3\,\text{kg m}^2\), and \(I_P\omega = 3 \times 6 = 18\,\text{kg m}^2\text{/s}\), the same as before. Its kinetic energy is \(\tfrac12 I_P\omega^2 = 54\,\text{J}\), which equals \(\tfrac12 Mv^2 + \tfrac12 I_{cm}\omega^2 = 36 + 18\).
The previous lesson used this same sum, \(MvR + I_{cm}\omega\), about a point on the ground line. There the body was skidding, so \(v\) and \(\omega R\) were not equal, and the sum could not be collapsed into \(I_P\omega\). The two-term form is the general one; \(I_P\omega\) is its shortcut for pure rolling.
A ring of mass \(2\,\text{kg}\) and radius \(0.4\,\text{m}\) rolls without slipping at \(2\,\text{m/s}\). Find its angular momentum about the point of the ground it touches, in kg m²/s.
A student claims that a rolling wheel has zero angular momentum about its contact point, because that point is at rest. What is wrong?
Choose the point that silences the unknown force
The torque law \(\vec\tau_{ext} = d\vec L/dt\) holds about any fixed point. So angular momentum is conserved about any point about which the outside torques cancel, whether or not it is conserved about other points. This turns into a method. When an unknown force acts, take moments about a point on its line of action. Its torque there is zero, however large the force.
The pivot of a hinged body was such a point in the chapter on Angular Momentum. A point on the ground line was such a point for the skidding ball. For a wheel that strikes a kerb, it is the edge of the kerb. The edge pushes with a large, brief force of unknown size and direction, but that force acts at the edge. During the brief impact the weight and the road's normal force give negligible angular impulse. The road's push also stops as soon as the wheel lifts.
A uniform disc of mass \(2\,\text{kg}\) and radius \(0.5\,\text{m}\) rolls without slipping at \(3\,\text{m/s}\) and strikes the edge of a kerb \(0.2\,\text{m}\) high. It does not bounce back or slide on the edge. Take \(g = 10\,\text{m/s}^2\). Find its angular velocity just after the impact, and decide whether it climbs onto the kerb.
Take moments about the edge \(E\). Before the impact, \(\omega = 3/0.5 = 6\,\text{rad/s}\) and \(I_{cm} = \tfrac12 \times 2 \times 0.25 = 0.25\,\text{kg m}^2\). The centre moves along a line \(d = 0.5 - 0.2 = 0.3\,\text{m}\) above \(E\).
Before: \(L_E = I_{cm}\omega + Mvd = 0.25 \times 6 + 2 \times 3 \times 0.3 = 3.3\,\text{kg m}^2\text{/s}\). After, the disc turns about \(E\), with \(I_E = I_{cm} + MR^2 = 0.75\,\text{kg m}^2\). Then \(3.3 = 0.75\,\omega'\).
\(\omega' = 4.4\,\text{rad/s}\), and the centre moves at \(4.4 \times 0.5 = 2.2\,\text{m/s}\). The kinetic energy is \(\tfrac12 \times 0.75 \times 4.4^2 \approx 7.3\,\text{J}\).
Climbing lifts the centre by \(0.2\,\text{m}\), which needs \(Mgh = 4\,\text{J}\). The disc has \(7.3\,\text{J}\), so it climbs, and its centre passes over the edge at about \(1.5\,\text{m/s}\).
The same working gives the least speed for climbing. The disc climbs only if \(\tfrac12 I_E\omega'^2 \ge Mgh\), which for this disc and kerb needs \(v\) of at least about \(2.23\,\text{m/s}\). The unknown force of the edge never entered. That is the whole point of choosing the point: the equation keeps the quantities you know and drops the one you do not.
A rolling wheel strikes the edge of a kerb and pivots onto it. Through the brief impact, which quantity of the wheel is conserved?
Now the trolley bag of the opening is explained. The jolt of the platform edge is unknown, but it acts at the edge, so angular momentum about the edge is kept. From it follows the spin of the wheel just after the jolt. If the energy of that spin is enough to lift the wheel, the bag hops up; if not, the wheel stops against the edge. A faster pull or a larger wheel makes the hop easier.
- Writing \(L = I_{cm}\omega\) for a body whose centre moves.Add the orbital term \(Mv_{cm}d\) about the chosen point.
- Adding the orbital and spin terms as plain numbers.Give each a sense, clockwise or anticlockwise, and add with signs.
- Conserving angular momentum about the centre of a wheel that strikes a kerb.The edge's force has a torque about the centre. Take moments about the edge.