Every batsman knows two kinds of shot. Strike the ball on the right part of the bat, and it flies away while the handle stays quiet in the hands. Strike it too near the handle or too near the toe, and the ball goes nowhere while the hands sting. The ball was the same, and so was the swing. Only the point of impact changed, and with it the jolt at the hands.
Three unknowns, three laws
Lay a uniform rod flat on smooth ice. A puck slides across the ice and strikes it at right angles, away from its centre. Just after the impact three things are unknown. They are the velocity \(v'\) of the puck, the velocity \(v_{cm}\) of the rod's centre, and the rod's angular velocity \(\omega\). A rod held by nothing is a free body, and three unknowns need three equations.
The ice is smooth, so no outside force acts along it. Linear momentum is conserved. With no outside force there is no outside torque either, so angular momentum is conserved about every fixed point. The easiest point is the one where the rod's centre is at the instant of impact; the rod's orbital term is then zero. If the impact is elastic, kinetic energy is conserved too. That makes three equations.
A uniform rod of mass \(M = 2\,\text{kg}\) and length \(1.2\,\text{m}\) lies at rest on smooth ice. A puck of mass \(m = 0.5\,\text{kg}\) slides at \(v = 6\,\text{m/s}\) and strikes one end at right angles. The impact is elastic. Find the velocity of the puck, the velocity of the rod's centre and the angular velocity of the rod just after.
The arm of the puck about the rod's centre is \(r = 0.6\,\text{m}\), and \(I_{cm} = \tfrac1{12}ML^2 = \tfrac1{12} \times 2 \times 1.44 = 0.24\,\text{kg m}^2\). Measure every velocity along the puck's first direction.
Momentum: \(0.5 \times 6 = 0.5v' + 2v_{cm}\). Angular momentum: \(0.5 \times 6 \times 0.6 = 0.5v' \times 0.6 + 0.24\,\omega\). Energy: \(\tfrac12 \times 0.5 \times 36 = \tfrac12 \times 0.5v'^2 + \tfrac12 \times 2v_{cm}^2 + \tfrac12 \times 0.24\,\omega^2\).
Write \(u = 6 - v'\). The first two give \(v_{cm} = u/4\) and \(\omega = 1.25u\). Putting these into the energy equation leaves \(u = 6\). So \(v' = 0\), \(v_{cm} = 1.5\,\text{m/s}\) and \(\omega = 7.5\,\text{rad/s}\).
The puck stops dead. The rod moves off at \(1.5\,\text{m/s}\) and turns at \(7.5\,\text{rad/s}\), sharing the \(9\,\text{J}\) as \(2.25\,\text{J}\) of translation and \(6.75\,\text{J}\) of rotation.
A check that uses none of this working is worth having. In an elastic impact the two bodies separate at the speed at which they approached, as the chapter on Collisions showed. The puck approached at \(6\,\text{m/s}\). Just after, the struck end of the rod moves at \(v_{cm} + \omega r = 1.5 + 7.5 \times 0.6 = 6\,\text{m/s}\), and the puck is at rest. They separate at \(6\,\text{m/s}\), exactly as they approached.
A particle strikes a free rod on smooth ice, and the impact is elastic. Which set of laws gives exactly the equations needed?
If the particle sticks
Now let the puck carry a lump of wet clay, so that it sticks to the end of the rod. The momentum and angular momentum equations stand exactly as before, because neither depended on the impact being elastic. The energy equation goes, because a sticking impact always loses kinetic energy. In its place comes the condition that the puck moves with the point it sticks to: \(v' = v_{cm} + \omega r\).
For the rod and the puck of Fig 6.1 the three equations are \(3 = 0.5v' + 2v_{cm}\), \(1.8 = 0.3v' + 0.24\,\omega\) and \(v' = v_{cm} + 0.6\,\omega\). They give \(v_{cm} = 0.75\,\text{m/s}\), \(\omega = 3.75\,\text{rad/s}\), and \(v' = 3\,\text{m/s}\) for the puck and the end together. The kinetic energy falls from \(9\,\text{J}\) to \(4.5\,\text{J}\); half is lost. As a check, the centre of mass of puck and rod moves at \(3/2.5 = 1.2\,\text{m/s}\), as momentum demands.
A particle strikes a free rod on ice and sticks to it. Which equation takes the place of the kinetic energy equation?
The point where the hinge feels nothing
Look again at the elastic strike in Fig 6.1. Just after the impact, one point of the rod is at rest: the point \(0.2\,\text{m}\) from the centre towards the far end. A free rod struck at a distance \(b\) from its centre starts to turn about a point \(k^2/b\) on the other side. Here \(k\) is its radius of gyration. If a hinge were placed exactly at that point, the hinge would not need to push at all.
Now hinge a uniform rod of mass \(M\) and length \(L\) at its top end. Strike it with a horizontal impulse \(P\) at a distance \(x\) below the hinge. During the blow the hinge may give an impulse \(J\) of its own, which is unknown. Take moments about the hinge, as the chapter on Angular Momentum did, and \(J\) drops out: \(Px = \tfrac13 ML^2\omega\). The linear law for the whole rod includes \(J\): \(P + J = Mv_{cm} = M\omega L/2\).
\[ \omega = \dfrac{3Px}{ML^2} \]
Why. the angular impulse about the hinge, \(Px\), equals \(I_{hinge}\,\omega = \tfrac13 ML^2\omega\).
\[ Mv_{cm} = M\omega\,\dfrac{L}{2} = \dfrac{3Px}{2L} \]
Why. the centre is \(L/2\) from the hinge, so it moves at \(\omega L/2\).
\[ J = Mv_{cm} - P \]
Why. the total impulse on the rod, \(P + J\), equals its change of momentum.
\[ J = P\left(\dfrac{3x}{2L} - 1\right) \]
Read it. put step 2 into step 3.
Slide the blow down the rod. Struck at \(0.25\,\text{m}\), the hinge must push back with about \(1.75\,\text{N s}\), against the blow; the rod pushes the hinge forward with the same impulse. Struck at the bottom, \(x = 0.9\,\text{m}\), the hinge must pull the rod forward with \(1.5\,\text{N s}\). Between the two, at \(x = 0.6\,\text{m}\), the impulse of the hinge is exactly zero. That point, \(2L/3\) from the hinge, is the centre of percussion.
The strength of the blow has cancelled. The centre of percussion is fixed by the shape of the body alone. A harder blow makes a larger jolt everywhere else, but at \(2L/3\) the jolt stays zero. Put \(J = 0\) in the two equations for any body, and the same working gives \(x_p = I_{hinge}/(Md)\). For a uniform rod that is \(\tfrac13 ML^2/(ML/2) = 2L/3\).
A batsman hits the ball harder than before, at the same point of the bat. What happens to the position of the bat's sweet spot?
A hanging rod is struck close to its hinge. Which way does the hinge push on the rod during the blow?
A uniform door \(1.2\,\text{m}\) wide swings on its hinges. Where should a door stopper be fixed, measured from the hinges, so that a slamming door gives no jolt to its hinges? Answer in metres.
A batsman's hands hold the handle, which acts like a hinge during the brief impact. A ball struck at the bat's centre of percussion leaves the handle with no jolt. The hands feel nothing, and none of the blow is wasted on them. A real bat is not a uniform rod, because its blade is thick and its handle thin. So its centre of percussion lies at \(I_{hand}/(Md)\) for its own shape, in the lower half of the blade. Bat makers shape the blade to put the sweet spot where the ball usually lands.
The dictionary, complete
The chapter on Dynamics of Rotation began a dictionary between straight-line motion and rotation. Every row is now earned. Take a law of motion along a line. Replace \(x, v, a, m, F, p\) by \(\theta, \omega, \alpha, I, \tau, L\), and the result is a true law of rotation. Rolling ties the two columns together through \(v = \omega R\) and \(a = \alpha R\).
Two facts do not translate, and this chapter has shown why they matter. A mass is a mass, but a moment of inertia must name its axis. A uniform rod has \(\tfrac1{12}ML^2\) about its centre and \(\tfrac13 ML^2\) about its end, and this chapter used both. A force is a force, but a torque and an angular momentum must name their point. The rolling disc of the previous lesson had \(18\) about the contact and \(6\) about its centre. Choosing the point is often the whole method.
Everything in this unit has been in a plane, with \(\vec\omega\) and \(\vec L\) along one fixed direction. A spinning top or a gyroscope turns its axis as well, and then \(\vec L\) need not point along \(\vec\omega\). The torque of gravity then swings the axis round slowly instead of toppling the top. That is precession, and it needs the full vector form of \(\vec\tau = d\vec L/dt\), beyond this course.
Now the batsman's sting is explained. The hands on the handle act as a hinge. Struck at the centre of percussion, the bat turns about the hands of its own accord, and the hands need give no impulse at all. Struck nearer the handle or nearer the toe, the hands must push back or pull, and that impulse is the sting. The sweet spot is a piece of geometry, the same for a gentle push and a full drive.
- Conserving linear momentum for a hinged body during an impact.A hinge gives an outside impulse. Only angular momentum about the hinge survives.
- Using kinetic energy when a particle sticks.Sticking loses energy. Use the contact condition \(v' = v_{cm} + \omega r\) instead.
- Expecting the sweet spot to move with the strength of the blow.\(x_p = I_{hinge}/(Md)\) contains no \(P\); it is fixed by the shape of the body.