Irodov 1.282: Angular Momentum Refuses the Rotation Axis

Problem Irodov 1.282

A thin uniform rod $AB$ of mass $m$ and length $l$ is clamped by its own midpoint $C$ to a straight vertical axle $OO'$ , whose two ends turn in fixed bearings. The clamp is rigid, so the rod pierces the axle at $C$ and holds a constant angle $\theta$ between itself and the axle. The whole assembly turns about $OO'$ without friction at a constant angular velocity $\omega$ . Find:

the magnitude and the direction of the rod’s angular momentum $\vect{M}$ about the point $C$ , and also the rod’s angular momentum about the rotation axis
the magnitude $|\Delta\vect{M}|$ of the change suffered by the vector $\vect{M}$ about $C$ during one half turn of the assembly
the magnitude $N$ of the moment of the external forces acting on the axle $OO'$ while the assembly turns

The arrangement. The axle OO' runs in fixed bearings at O and O' and turns at ω; the rod AB is clamped by its midpoint C to the axle, holding the fixed angle θ between the rod and the axle. — The arrangement. The axle $OO'$ runs in fixed bearings at $O$ and $O'$ and turns at $\vect\omega$ ; the rod $AB$ is clamped by its midpoint $C$ to the axle, holding the fixed angle $\theta$ between the rod and the axle.

Given

$m$
$l$
$\theta$
$\omega$

Find

(a)

M

about

C

, its direction, and

M_z

about the axle; (b)

|\Delta\vect M|

over a half turn; (c)

N

Solution

Step 1A frame that turns with the rod

Put the origin at $C$ and let $\vect k$ point up along the axle. At the instant drawn, let $\vect i$ be the horizontal unit vector in the plane that holds the axle and the rod, pointing to the side the end $B$ leans toward, and let $\vect j$ complete the right-handed triad. The rotation and the rod’s own direction are then both fixed by $\theta$ .

\vect\omega = \omega\vect k, \qquad \vect e = \sin\theta\,\vect i+\cos\theta\,\vect k, \qquad \vect r = s\,\vect e \tag{1}

Here $s$ is the signed distance of an element from $C$ , measured along the rod and counted positive toward $B$ ; the rod being uniform, its linear density is constant and the element’s mass follows.

dm = \frac{m}{l}\,ds, \qquad -\frac{l}{2}\le s\le \frac{l}{2}

One element of the rod: a mass dm a distance s from C along the rod, and so a perpendicular distance sθ from the axle. It moves into the page as the axle turns at ω, and contributes d M about C. — One element of the rod: a mass $dm$ a distance $s$ from $C$ along the rod, and so a perpendicular distance $s\sin\theta$ from the axle. It moves into the page as the axle turns at $\vect\omega$ , and contributes $d\vect M$ about $C$ .

Step 2The velocity of one element

Every point of a rigid body turning about a fixed axis moves with $\vect v=\vect\omega\times\vect r$ . Expanding that product in the basis of (1) collapses it onto a single direction.

\begin{aligned} \vect v &= \vect\omega\times\vect r \\ &= \omega s\,(\vect k\times\vect e) \\ &= \omega s\bigl(\sin\theta\,(\vect k\times\vect i)+\cos\theta\,(\vect k\times\vect k)\bigr) \\ &= \omega s\sin\theta\,\vect j \end{aligned} \tag{2}

So the element travels perpendicular to the plane of the rod and the axle, at the speed $\omega|s|\sin\theta$ , which is $\omega$ times its distance from the axle.

Step 3The angular momentum one element carries

The angular momentum of the element about $C$ is $\vect r\times\vect v\,dm$ . Substituting the velocity of (2) leaves one more cross product, between the rod’s direction and $\vect j$ .

\begin{align*}d\vect M &= \vect r\times\vect v\,dm \\ &= (s\vect e)\times(\omega s\sin\theta\,\vect j)\,dm \\ &= \omega s^2\sin\theta\,(\vect e\times\vect j)\,dm\end{align*}

That remaining product is fixed by the triad alone.

\vect e\times\vect j = \sin\theta\,(\vect i\times\vect j)+\cos\theta\,(\vect k\times\vect j) = \sin\theta\,\vect k-\cos\theta\,\vect i

d\vect M = \frac{m\omega\sin\theta}{l}\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr)s^2\,ds

(3)

Step 4Summing the rod into one vector

The direction in (3) does not depend on $s$ , so it comes out of the integral and only $s^2$ is left to integrate along the rod.

\begin{align*}\vect M &= \frac{m\omega\sin\theta}{l}\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr)\int_{-l/2}^{+l/2}s^2\,ds \\ &= \frac{m\omega\sin\theta}{l}\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr)\left.\frac{s^3}{3}\right|_{-l/2}^{+l/2} \\ &= \frac{m\omega\sin\theta}{l}\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr)\frac{l^3}{12}\end{align*}

\vect M = \frac{m\omega l^2}{12}\sin\theta\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr)

(4)

The angular momentum about C at one instant. The gray dashed ray is the axle direction below C; θ is marked between it and the rod, 90°-θ between the axle above C and M. The dashed arrows are M_z and M_⊥. — The angular momentum about $C$ at one instant. The gray dashed ray is the axle direction below $C$ ; $\theta$ is marked between it and the rod, $90^\circ-\theta$ between the axle above $C$ and $\vect M$ . The dashed arrows are $M_z$ and $M_\perp$ .

Step 5Magnitude and direction of

\vect M

The bracket in (4) is a unit vector, since its two components are a sine and a cosine of the same angle.

\bigl|\sin\theta\,\vect k-\cos\theta\,\vect i\bigr| = \sqrt{\sin^2\theta+\cos^2\theta} = 1

M = \frac{1}{12}\,m\omega l^2\sin\theta

(5)

For the direction, test $\vect M$ against the rod’s own direction $\vect e$ of (1).

\vect M\cdot\vect e = \frac{m\omega l^2}{12}\sin\theta\bigl(\sin\theta\cos\theta-\cos\theta\sin\theta\bigr) = 0

So $\vect M$ is perpendicular to the rod, and since it has no $\vect j$ component it lies in the plane that holds the rod and the axle. Its tilt away from the axle follows from the same bracket.

\cos\angle(\vect M,\vect k) = \frac{\vect M\cdot\vect k}{M} = \frac{(m\omega l^2/12)\sin^2\theta}{(m\omega l^2/12)\sin\theta} = \sin\theta

Perpendicular to the rod, at

90^\circ-\theta

to the axle.

As the assembly turns, $\vect M$ keeps that tilt and sweeps out a cone about the axle.

Step 6The component along the rotation axis

The angular momentum about the rotation axis is the projection of $\vect M$ onto the axle direction, so it is one dot product away from (4).

\begin{align*}M_z &= \vect M\cdot\vect k \\ &= \frac{m\omega l^2}{12}\sin\theta\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr)\cdot\vect k\end{align*}

M_z = \frac{1}{12}\,m\omega l^2\sin^2\theta = M\sin\theta

(6)

The rod and its angular momentum at one instant (solid) and half a turn later (dashed). The two tips sit at the same height, and the increment Δ M that joins them is horizontal. — The rod and its angular momentum at one instant (solid) and half a turn later (dashed). The two tips sit at the same height, and the increment $\Delta\vect M$ that joins them is horizontal.

Step 7The change in

\vect M

over a half turn

A half turn about the axle reverses the transverse direction $\vect i$ and leaves the axial direction $\vect k$ alone, which gives the later vector at once; subtracting (4) gives the increment.

\begin{align*}\vect M' &= \frac{m\omega l^2}{12}\sin\theta\bigl(\sin\theta\,\vect k+\cos\theta\,\vect i\bigr) \\ \Delta\vect M &= \vect M'-\vect M \\ &= \frac{m\omega l^2}{12}\sin\theta\cdot 2\cos\theta\,\vect i\end{align*}

Folding the product of the two trigonometric factors into the double angle gives the modulus.

|\Delta\vect M| = \frac{1}{12}\,m\omega l^2\sin 2\theta

(7)

Step 8The couple the bearings must supply

The moment of the external forces about $C$ is the rate of change of $\vect M$ . Gravity contributes nothing here, since it acts at the rod’s center of mass, which is $C$ itself. In the laboratory frame $\vect M$ keeps both its length and its tilt and simply turns bodily with the rod, so its derivative is the same rotation formula that (2) applied to a position vector.

\begin{align*}\vect N &= \frac{d\vect M}{dt} = \vect\omega\times\vect M \\ &= \omega\vect k\times\frac{m\omega l^2}{12}\sin\theta\bigl(\sin\theta\,\vect k-\cos\theta\,\vect i\bigr) \\ &= -\frac{m\omega^2l^2}{12}\sin\theta\cos\theta\,(\vect k\times\vect i)\end{align*}

With $\vect k\times\vect i=\vect j$ , this moment is horizontal and perpendicular to the plane holding the rod and the axle, and it turns with them. Folding the trigonometric product into the double angle again gives its magnitude.

N = \frac{1}{24}\,m\omega^2l^2\sin 2\theta

(8)

Check

Dimensionally $m\omega l^2$ is a mass times a frequency times an area, which is an angular momentum, and $m\omega^2l^2$ is a mass times an acceleration times a length, which is a moment. The two degenerate tilts test the vector structure rather than the algebra. At $\theta=0$ the rod lies along the axle, every element sits on the axis, and all four boxed quantities vanish. At $\theta=90^\circ$ the rod is perpendicular to the axle, $\vect M$ lies along the axle so that $M_z=M$ , and both the increment and the moment vanish because $\vect M$ then has nowhere to swing. Parts (b) and (c) must also agree through the moment’s impulse: a moment of fixed magnitude $N$ whose direction turns uniformly with the shaft delivers $\bigl|\int\vect N\,dt\bigr|=2N/\omega$ over a half turn, and $2N/\omega$ is exactly the boxed $|\Delta\vect M|$ .

Remark

Resolving the rotation instead of the rod makes the whole answer immediate. Split $\vect\omega$ into a part along the rod, of size $\omega\cos\theta$ , and a part perpendicular to it, of size $\omega\sin\theta$ . A thin rod has no moment of inertia about its own length, so the along-the-rod part of $\vect\omega$ carries no angular momentum whatever; the perpendicular part meets the moment of inertia $ml^2/12$ and yields an angular momentum along itself, perpendicular to the rod and of size $(ml^2/12)\,\omega\sin\theta$ , which is the magnitude and the direction boxed in part (a) at once; projecting that onto the axle costs one more factor of $\sin\theta$ and gives the third box, $M_z$ . That is the general lesson: $\vect M$ is parallel to $\vect\omega$ only when the body turns about one of its own principal axes, and the principal axes of a thin rod at its center are the rod itself and every direction perpendicular to it. The axle coincides with one of them only at $\theta=0$ and $\theta=90^\circ$ , which is precisely where the transverse part of $\vect M$ , the half-turn increment and the bearing couple all vanish together.