Irodov 1.273: What the Halves Owe Each Other – Amit & Deepali Sinha Foundation

Problem Irodov 1.273

A uniform rod of mass $m = 5.0$ kg and length $l = 90$ cm lies on a smooth (frictionless) horizontal surface. One end of the rod is struck by a horizontal impulse $J = 3.0$ N $\cdot$ s directed perpendicular to the rod, and as a result the rod acquires linear momentum $p = 3.0$ N $\cdot$ s. Determine the internal force each half of the rod exerts on the other while this subsequent motion continues.

The rod on the smooth surface, seen from above. It lies free, of mass m and length l, with its center of mass marked, and the impulse J is delivered at the end A at right angles to the rod. — The rod on the smooth surface, seen from above. It lies free, of mass $m$ and length $l$ , with its center of mass marked, and the impulse $\vect J$ is delivered at the end $A$ at right angles to the rod.

Given

$m = 5.0$ kg
$l = 90$ cm
$J = 3.0$ N $\cdot$ s at one end, perpendicular
$p = 3.0$ N $\cdot$ s
smooth surface

Find

Find the force

F

between the two halves during the subsequent motion

Solution

Step 1What the blow leaves behind

The blow is a horizontal impulse applied perpendicular to the rod at one end. It hands the rod linear momentum $p = J$ , which is what the problem states and which is the whole of the linear bookkeeping: the surface is smooth, so nothing else pushes horizontally. The center of mass therefore leaves with

v_c = \frac{J}{m} .

The angular bookkeeping is separate. About the center of mass the impulse has the arm $l/2$ , so the angular impulse it delivers is $J\,l/2$ , and with $I = ml^{2}/12$ for a uniform rod about its center,

J\,\frac{l}{2} = I\omega = \frac{ml^{2}}{12}\,\omega .

\omega = \frac{6J}{ml}

(1)

Step 2After the blow nothing pushes the rod at all

The rod once the blow is over, seen from above. Its center moves off with v_c in the direction of the blow, and the rod turns about that center with the constant ω. — The rod once the blow is over, seen from above. Its center moves off with $\vect v_c$ in the direction of the blow, and the rod turns about that center with the constant $\vect\omega$ .

Once the blow is over, the rod slides on a smooth horizontal surface. Gravity is balanced by the normal reaction, and there is no horizontal external force. So the center of mass moves in a straight line at constant speed, and, no external moment acting, the rod turns about that center at the constant rate $\omega$ found above. The motion is uniform translation plus uniform rotation.

Step 3Isolate one half

Cut the rod at its center and take the half between the cut and the end $B$ . Its mass is $m/2$ and its own center of mass is at the midpoint of that half, a distance $l/4$ from the cut.

One half of the rod. Its center of mass runs round a circle of radius l/4 about the rod's center at the constant rate ω, so its acceleration is centripetal, and only the other half can supply it. — One half of the rod. Its center of mass runs round a circle of radius $l/4$ about the rod’s center at the constant rate $\omega$ , so its acceleration is centripetal, and only the other half can supply it.

That point is carried round the rod’s center of mass at the constant angular rate $\omega$ , on a circle of radius $l/4$ . The rod’s center of mass itself has no acceleration, so the acceleration of the half’s center of mass is exactly the centripetal acceleration of that circular motion,

a = \omega^{2}\,\frac{l}{4},

directed along the rod toward the center.

Step 4Newton for the half

The only horizontal force on the half is what the other half exerts on it across the cut, since the surface is smooth. Newton’s second law for the half is therefore

F = \frac{m}{2}\,a = \frac{m}{2}\cdot\omega^{2}\frac{l}{4} = \frac{m\omega^{2}l}{8} .

Substituting $\omega$ from (1) and squaring it,

F = \frac{ml}{8}\cdot\frac{36J^{2}}{m^{2}l^{2}} .

F = \frac{9}{2}\,\frac{J^{2}}{ml}

(2)

Putting in the numbers, with $J^{2} = 9.0\ \mathrm{N^{2}s^{2}}$ , evaluate $ml$ first and then $F$ :

\begin{aligned} ml &= 5.0 \times 0.90 = 4.5\ \mathrm{kg\,m}, \\ F &= \frac{9}{2}\cdot\frac{9.0}{4.5} = 9.0\ \mathrm{N}. \end{aligned}

Check

The units work out: $J^{2}/(ml)$ is $(\mathrm{N\,s})^{2}/(\mathrm{kg\,m}) = \mathrm{N}\cdot(\mathrm{kg\,m\,s^{-2}})(\mathrm{s^{2}})/(\mathrm{kg\,m}) = \mathrm{N}$ . And the given $p$ is a real check rather than spare data: $p = mv_c = m(J/m) = J = 3.0$ N $\cdot$ s, exactly the stated value, which confirms that the whole of the impulse went into the momentum of the rod and none of it into the surface.

Remark

The force is a tension: each half pulls the other inwards, toward the center, since that is the direction each half’s acceleration points. Its value is fixed once and for all by the blow and does not change with time, since $\omega$ is constant. Note also that it grows as $J^{2}$ , so doubling the blow quadruples the internal force.