Irodov 1.275: Where to Hit and Feel Nothing – Amit & Deepali Sinha Foundation

Problem Irodov 1.275

A uniform rod of mass $M$ and length $l$ hangs vertically from a horizontal hinge at its upper end, about which it can turn freely. A bullet of mass $m$ , flying horizontally, hits the rod’s lower end and embeds itself there, after which the rod swings up until it has turned through an angle $\alpha$ from the vertical. The bullet is much lighter than the rod, $m \ll M$ . Find:

the speed $v$ of the bullet before it strikes
the increment $\Delta p$ of the total linear momentum of the bullet and rod together over the duration of the impact, and what is responsible for that change
the distance $x$ from the hinge at which the bullet would have to strike for the total linear momentum of bullet and rod to be unchanged by the impact

Just before the impact: the rod M of length l hangs from its upper hinge, and the bullet m arrives horizontally at the rod's lower end with speed v. — Just before the impact: the rod $M$ of length $l$ hangs from its upper hinge, and the bullet $m$ arrives horizontally at the rod’s lower end with speed $v$ .

At the top of the swing: the rod has turned through α from its original vertical position (dashed), carrying the embedded bullet with it, and its center of mass has risen by h from the lower dot to the upper one. — At the top of the swing: the rod has turned through $\alpha$ from its original vertical position (dashed), carrying the embedded bullet with it, and its center of mass has risen by $h$ from the lower dot to the upper one.

Given

$M$
$l$
$m\ll M$
$\alpha$
$g$

Find

(a)

v

, (b)

\Delta p

and its cause, (c) the strike distance

x

that makes

\Delta p

vanish

Solution

Step 1What the blow conserves

The impact is short enough that the finite forces, gravity and any steady part of the hinge reaction, transfer negligible impulse during it. The hinge can deliver a large impulsive force, but its line of action passes through the hinge, so it contributes no angular impulse about that point. Angular momentum about the hinge is therefore conserved across the blow, while linear momentum in general is not. With $\omega$ the angular velocity just after the blow and $I$ the moment of inertia of the rod about its end, and dropping the bullet’s own $ml^2$ against $I$ because $m\ll M$ ,

I = \frac{Ml^2}{3}

mvl = I\omega = \frac{Ml^2}{3}\,\omega

(1)

Step 2How far the swing determines the launch rate

After the blow the rod turns freely, so its energy is conserved and the kinetic energy it starts with is spent raising its center of mass. That center lies at $l/2$ from the hinge, and the figure shows the rise as the vertical gap between its two positions. The bullet’s own contribution to both energy and weight is of relative order $m/M$ and is dropped.

\begin{align*}h &= \frac{l}{2}\bigl(1-\cos\alpha\bigr) \\ \frac{1}{2}I\omega^2 &= Mgh \\ \frac{1}{2}\cdot\frac{Ml^2}{3}\,\omega^2 &= Mg\,\frac{l}{2}\bigl(1-\cos\alpha\bigr) \\ \omega^2 &= \frac{3g}{l}\bigl(1-\cos\alpha\bigr)\end{align*}

The half-angle identity turns that into a form whose square root is clean.

1-\cos\alpha = 2\sin^2\frac{\alpha}{2}

\omega = \sqrt{\frac{6g}{l}}\;\sin\frac{\alpha}{2}

(2)

Step 3The bullet’s speed

Solving (1) for $v$ and substituting (2) answers part (a).

\begin{align*}v &= \frac{Ml}{3m}\,\omega \\ &= \frac{Ml}{3m}\sqrt{\frac{6g}{l}}\;\sin\frac{\alpha}{2} \\ &= \frac{M}{3m}\sqrt{6gl}\;\sin\frac{\alpha}{2}\end{align*}

v = \frac{M}{m}\sqrt{\frac{2gl}{3}}\;\sin\frac{\alpha}{2}

(3)

Step 4The momentum the hinge injects

Before the blow only the bullet moves, so the system’s momentum is $mv$ . Immediately after, the rod and the bullet turn together about the hinge; the rod’s own momentum is its mass times the speed of its center of mass, at $l/2$ from the hinge, and the bullet’s is smaller by the factor $m/M$ and is dropped. Writing both in terms of $\omega$ with the help of (1) lets the two be subtracted directly.

\begin{align*}p_{\mathrm{before}} &= mv = \frac{Ml}{3}\,\omega \\ p_{\mathrm{after}} &= M\cdot\frac{\omega l}{2} = \frac{Ml}{2}\,\omega \\ \Delta p &= \frac{Ml}{2}\,\omega - \frac{Ml}{3}\,\omega = \frac{Ml}{6}\,\omega\end{align*}

Substituting (2) gives the size of that increment.

\Delta p = \frac{Ml}{6}\sqrt{\frac{6g}{l}}\;\sin\frac{\alpha}{2} = M\sqrt{\frac{gl}{6}}\;\sin\frac{\alpha}{2}

(4)

The hinge supplies it, as a horizontal impulsive reaction forwards.

Part (c): the bullet m arrives with velocity v and strikes the rod M at P, a distance x below the hinge; the rod's center of mass C moves off at u immediately afterwards. — Part (c): the bullet $m$ arrives with velocity $\vect v$ and strikes the rod $M$ at $P$ , a distance $x$ below the hinge; the rod’s center of mass $C$ moves off at $\vect u$ immediately afterwards.

Step 5The strike point that spares the hinge

Repeat the momentum count of part (b) with the bullet arriving at a general distance $x$ from the hinge. Angular momentum about the hinge again fixes the launch rate, and the rod’s center of mass again moves off at $\omega l/2$ , so the two momenta can be set equal and the common factor $\omega$ removed.

\begin{align*}mvx &= I\omega \\ p_{\mathrm{before}} = mv &= \frac{I\omega}{x} \\ p_{\mathrm{after}} &= M\,\frac{\omega l}{2}\end{align*}

Equating the two and inserting $I=Ml^2/3$ leaves $x$ alone.

\begin{align*}\frac{I\omega}{x} &= \frac{M\omega l}{2} \\ x &= \frac{2I}{Ml} = \frac{2}{Ml}\cdot\frac{Ml^2}{3}\end{align*}

x = \frac{2}{3}\,l

(5)

Check

Every result is dimensionally sound: $\sqrt{gl}$ is a speed, so $v$ is a speed and $\Delta p$ a mass times a speed, and $x$ is a length. Both $v$ and $\Delta p$ carry the factor $\sin(\alpha/2)$ and so vanish as $\alpha\to0$ , where the rod barely moves; $x$ carries no $\alpha$ at all, being a property of the rod and its hinge rather than of the blow. The two parts also cross-check each other: part (b) found $\Delta p = (Ml/6)\omega$ while $mv=(Ml/3)\omega$ , so the hinge’s impulse is exactly half the bullet’s own momentum, and substituting the boxed $v$ into $mv/2$ reproduces the boxed $\Delta p$ term for term. Finally, part (c) is the root of the very expression part (b) evaluated: at $x=2l/3$ the term $I\omega/x$ reduces to $M\omega l/2$ , the two momenta coincide, and the increment vanishes as it should.

Remark

The point at $2l/3$ is the center of percussion of the rod for this hinge, the “sweet spot” a bat or racket has: a blow delivered there sets the body turning without jarring the pivot, while a blow anywhere else leaves a residue the pivot has to absorb, forwards for a strike below the sweet spot and backwards for one above it. In general the center of percussion sits at $I/Md$ from the pivot, with $d$ the pivot to center of mass distance, and the same combination is the equivalent length of the body as a physical pendulum, so a rod hinged at its end swings with the period of a simple pendulum of length $2l/3$ . The relation is also reciprocal: hang the rod instead from a hinge at $2l/3$ and its center of percussion moves to the old pivot, which is why the two points are called conjugate.