At a village mela in Punjab, one stall has a game with a wooden paddle hanging from a pin. You throw a ball of wet clay at the bottom of the paddle. The clay sticks, and the paddle swings up. If it swings past a painted mark, you win a prize. The stall keeper knows that a hard throw is not enough. Where the clay hits, and how heavy the paddle is, matter just as much.
What the pivot does
Let a ball strike a rod that hangs from a fixed pivot. The pivot must keep its own point still. So during the impact it pushes on the rod with a force that is large, brief and unknown, an impulsive force. That force is external to the system of ball and rod. It changes their total linear momentum, so linear momentum is not conserved through the impact.
But the pivot's force acts at the pivot. Its moment arm about the pivot is zero, so its torque about the pivot is zero, however large the force is. Gravity does act about the pivot, but it is an ordinary force. Over the tiny duration of the impact, its angular impulse is negligible. So the angular momentum of the system about the pivot is the same just before and just after the impact.
Through a brief impact on a body held by a fixed pivot, the angular momentum of the whole system about the pivot is conserved. The linear momentum is not, because the pivot's force is external.
Every such problem is solved the same way, in four steps.
- Find the angular momentum about the pivot just before, \(L_i\). For an incoming particle it is \(mvd\), with \(d\) the moment arm from the pivot to its line of motion.
- Find the total moment of inertia \(I_{\text{total}}\) about the pivot just after, with every part taken about that same point.
- Set \(L_i = I_{\text{total}}\,\omega\) and solve for \(\omega\).
- If the question asks about energy, compare the kinetic energies. After the impact, energy conservation takes over for the swing.
A ball strikes the free end of a rod hinged at the ceiling and sticks to it. Through the instant of the impact, which quantity of the ball-and-rod system is conserved?
A uniform rod of mass \(M = 1.2\,\text{kg}\) and length \(\ell = 1.0\,\text{m}\) hangs from a hinge at its top end. A ball of mass \(m = 0.2\,\text{kg}\), moving horizontally at \(v = 6\,\text{m/s}\), strikes the lower end and sticks. Take \(g = 10\,\text{m/s}^2\). Find the angular velocity just after the impact and the greatest angle of the swing.
Take moments about the hinge. Just before, only the ball moves, with moment arm \(\ell\): \(L_i = mv\ell = 0.2 \times 6 \times 1.0 = 1.2\,\text{kg m}^2\text{/s}\).
Just after, rod and ball turn together: \(I_{\text{total}} = \tfrac13 M\ell^2 + m\ell^2 = 0.4 + 0.2 = 0.6\,\text{kg m}^2\). Then \(L_i = I_{\text{total}}\,\omega\).
\(\omega = 1.2/0.6 = 2\,\text{rad/s}\). The kinetic energy just after is \(\tfrac12 \times 0.6 \times 2^2 = 1.2\,\text{J}\). In the swing, the rod's centre and the ball rise by \(\tfrac12\ell(1 - \cos\theta)\) and \(\ell(1 - \cos\theta)\), so \((Mg\ell/2 + mg\ell)(1 - \cos\theta) = 8(1 - \cos\theta) = 1.2\).
\(\omega = 2\,\text{rad/s}\) just after. Then \(\cos\theta = 0.85\), so the rod swings up to \(\theta \approx 31.8°\).
The ball brought \(\tfrac12 \times 0.2 \times 6^2 = 3.6\,\text{J}\), and only \(1.2\,\text{J}\) survived the impact. Two thirds of the kinetic energy went into squashing the ball as it stuck. Angular momentum was conserved; kinetic energy was not.
The hinge's impulse can even be measured. Just after the impact, the rod's centre moves at \(\omega\ell/2 = 1\,\text{m/s}\) and the ball at \(\omega\ell = 2\,\text{m/s}\). So the total momentum is \(1.2 \times 1 + 0.2 \times 2 = 1.6\,\text{kg m/s}\). Just before, it was \(0.2 \times 6 = 1.2\,\text{kg m/s}\). The system gained \(0.4\,\text{N s}\) in the direction of the ball's motion. The hinge supplied it, which is exactly why linear momentum could not be used.
A uniform rod of mass \(0.9\,\text{kg}\) and length \(1.2\,\text{m}\) hangs from a hinge at its top. A \(0.1\,\text{kg}\) ball moving horizontally at \(9\,\text{m/s}\) strikes its lower end and sticks. Find the angular velocity just after the impact, in rad/s.
A disc on an axle, and a rod caught by a peg
A uniform disc of mass \(4\,\text{kg}\) and radius \(0.5\,\text{m}\) turns freely on a fixed central axle and is at rest. A \(0.5\,\text{kg}\) lump of putty moving at \(6\,\text{m/s}\) along a tangent strikes the rim and sticks. Its moment arm about the axle is the radius, so \(L_i = 0.5 \times 6 \times 0.5 = 1.5\,\text{kg m}^2\text{/s}\). After the impact, \(I_{\text{total}} = \tfrac12 \times 4 \times 0.5^2 + 0.5 \times 0.5^2 = 0.625\,\text{kg m}^2\). So \(\omega = 1.5/0.625 = 2.4\,\text{rad/s}\).
Aim the same putty at the axle instead. Its line of motion now passes through the pivot, so its moment arm is zero and \(L_i = 0\). The disc does not turn at all, however fast the putty moves. The axle absorbs the whole impact. What sets a pivoted body turning is the moment arm of the incoming momentum, not its size alone.
Two identical lumps of putty, at the same speed, strike a pivoted disc at rest and stick at its rim. One moves along a tangent to the rim; the other moves straight towards the axle. Which sets the disc turning faster?
A rod can also start to turn without any ball. A uniform rod of mass \(M = 3\,\text{kg}\) and length \(\ell = 1.2\,\text{m}\) slides sideways across a smooth table at \(v = 4\,\text{m/s}\). It stays at right angles to its motion. Then one end catches on a fixed peg, which becomes a pivot. Just before, the rod translates, and its angular momentum about the peg is that of its centre of mass, at moment arm \(\ell/2\).
So \(L_i = Mv\ell/2 = 3 \times 4 \times 0.6 = 7.2\,\text{kg m}^2\text{/s}\). About its end the rod has \(I = \tfrac13 M\ell^2 = 1.44\,\text{kg m}^2\), so \(\omega = 7.2/1.44 = 5\,\text{rad/s}\). In symbols, \(\omega = 3v/2\ell\), whatever the mass. The centre then moves at \(\omega\ell/2 = 3\,\text{m/s}\), slower than before, and the far end at \(6\,\text{m/s}\), faster than before.
A uniform rod of mass \(2\,\text{kg}\) and length \(1.0\,\text{m}\) slides sideways on a smooth table at \(6\,\text{m/s}\), at right angles to its length. One end catches on a fixed peg. Find the angular velocity of the rod just after, in rad/s.
How much energy a sticking impact keeps
The three impacts of this lesson all conserve angular momentum about the pivot, and all lose kinetic energy. But they lose very different shares. One ratio explains all three. Let a body of mass \(m\) arrive at speed \(v\). The line of motion of its centre of mass passes at moment arm \(d\) from the pivot. After it sticks, everything turns about the pivot with moment of inertia \(I_{\text{total}}\).
\[ L_i = mvd, \qquad K_i = \tfrac12 mv^2 \]
Why. before the impact the body moves in a straight line at moment arm \(d\) from the pivot.
\[ K_f = \dfrac{L_i^2}{2I_{\text{total}}} = \dfrac{m^2v^2d^2}{2I_{\text{total}}} \]
Why. after the impact \(L = L_i\), and from lesson 4 the kinetic energy of rotation is \(L^2/2I\).
\[ \dfrac{K_f}{K_i} = \dfrac{m^2v^2d^2}{2I_{\text{total}}}\cdot\dfrac{2}{mv^2} = \dfrac{md^2}{I_{\text{total}}} \]
Read it. divide, and \(m\), \(v^2\) and the factors of 2 cancel.
The numerator \(md^2\) is the moment of inertia that the incoming body alone would have about the pivot. So the fraction kept is the incoming body's share of the final moment of inertia. For the ball on the rod, \(0.2/0.6 = \tfrac13\). For the putty on the disc, \(0.125/0.625 = \tfrac15\). For the rod caught by the peg, the incoming body is the rod itself, with \(md^2 = 3 \times 0.6^2 = 1.08\), and \(1.08/1.44 = \tfrac34\). A light body stuck to a heavy one keeps little of its energy.
A uniform disc of mass \(9\,\text{kg}\) and radius \(0.4\,\text{m}\) is free to turn on a fixed central axle. A \(0.5\,\text{kg}\) lump of putty moving along a tangent strikes its rim and sticks. What percentage of the putty's kinetic energy survives the impact?
A lump of putty strikes the rim of a pivoted wheel and sticks. It is then thrown in the same way at a much heavier wheel of the same radius. Compared with the first impact, the fraction of kinetic energy that survives is
In 1742 the English engineer Benjamin Robins published New Principles of Gunnery. Nobody then could time a musket ball in flight. Robins fired the ball into a heavy block of wood faced with iron, hung from a pivot above it. The ball stuck, and the block swung up. A ribbon pulled out by the swing measured how far it went. From the swing Robins worked back to the speed just after the impact, and from that to the speed of the ball. The pivot made angular momentum about the pivot, not linear momentum, the quantity that survived.
Now return to the paddle at the mela. Suppose the paddle is like the rod of the figure, \(1\,\text{m}\) long and \(1.2\,\text{kg}\), and the clay ball is \(0.2\,\text{kg}\) at \(6\,\text{m/s}\). Hit at the bottom, the paddle starts at \(2\,\text{rad/s}\) and swings to about \(32°\). Hit halfway up, the moment arm is halved, and the paddle swings only to about \(20°\). A heavier paddle has a larger \(I_{\text{total}}\), so it keeps an even smaller share of the ball's energy. The stall keeper's prize depends on the moment arm and the moment of inertia, not on muscle alone.
- Using conservation of linear momentum in an impact on a pivoted body.The pivot's impulsive force is external. Conserve angular momentum about the pivot instead.
- Taking some parts of \(I_{\text{total}}\) about the centre of mass and others about the pivot.Every part must be about the pivot. Use \(\tfrac13 M\ell^2\) for a rod hinged at its end, and the parallel-axis theorem where needed.
- Using energy conservation through the impact itself.A sticking impact loses energy. Use angular momentum for the impact, and energy only for the swing afterwards.