Irodov 1.188: Which Point Keeps the Momentum Fixed – Amit & Deepali Sinha Foundation

Problem Irodov 1.188

A small ball of mass $m$ hangs from a fixed point $O$ on a thread of length $l$ and moves at constant angular velocity $\omega$ around a horizontal circle, the thread sweeping out a cone of constant half angle about the vertical through $O$ . Identify every point relative to which the ball’s angular momentum vector $\vect M$ stays exactly constant in time, and find the magnitude of the change in the angular momentum relative to $O$ that accumulates over half of one revolution.

The conical pendulum of the statement: a thread of length l from the fixed pivot O at constant angle θ to the vertical, the ball of mass m sweeping a horizontal circle of radius r at angular velocity ω. — The conical pendulum of the statement: a thread of length $l$ from the fixed pivot $O$ at constant angle $\theta$ to the vertical, the ball of mass $m$ sweeping a horizontal circle of radius $r$ at angular velocity $\omega$ .

Given

$m$
$l$
$\omega$

Find

the point(s) about which

\vect M

is constant;

|\Delta\vect M_O|

over half a revolution

Solution

Step 1The cone angle from the force balance

Left, the free body: tension T along the thread toward O, weight m g down, and the circle's center C a depth lθ below O. Right, the unit vectors z, r, φ at the ball, with φ increasing the way it travels. — Left, the free body: tension $\vect T$ along the thread toward $O$ , weight $m\vect g$ down, and the circle’s center $C$ a depth $l\cos\theta$ below $O$ . Right, the unit vectors $\hat{\vect z}$ , $\hat{\vect r}$ , $\hat{\vect\varphi}$ at the ball, with $\varphi$ increasing the way it travels.

The ball moves on a horizontal circle of radius $r=l\sin\theta$ at constant height, so its acceleration is purely centripetal, $\omega^2 r$ toward $C$ . Resolving tension $\vect T$ and weight $m\vect g$ vertically and horizontally,

\begin{align*}T\cos\theta &= mg \\ T\sin\theta &= m\omega^2 r = m\omega^2 l\sin\theta.\end{align*}

Dividing the second equation by $\sin\theta$ isolates the tension, and substituting that into the first eliminates it in favor of $\theta$ .

\begin{align*}T &= m\omega^2 l \\ m\omega^2 l\cos\theta &= mg\end{align*}

\cos\theta = \dfrac{g}{\omega^2 l}

(1)

Step 2Only the center of the circle keeps

\vect M

exactly constant

At every instant the ball’s net force is purely centripetal, $\vect F_{\text{net}}=-m\omega^2 r\,\hat{\vect r}$ , horizontal and directed from the ball toward the axis. For a general point $P$ , write its position as $P=C+\vect d+z_P\hat{\vect z}$ , where $\vect d$ is $P$ ’s horizontal offset from $C$ (possibly $\vect 0$ ) and $z_P$ its height above $C$ . The ball’s position relative to $P$ is $r\hat{\vect r}-\vect d-z_P\hat{\vect z}$ , so the torque about $P$ is

\begin{align*}\vect\tau_P &= (r\hat{\vect r}-\vect d-z_P\hat{\vect z})\times(-m\omega^2 r\,\hat{\vect r}) \\ &= m\omega^2 r\,(\vect d\times\hat{\vect r}) + z_P\, m\omega^2 r\,(\hat{\vect z}\times\hat{\vect r}) \\ &= m\omega^2 r\,(\vect d\times\hat{\vect r}) + z_P\, m\omega^2 r\,\hat{\vect\varphi},\end{align*}

using the cylindrical identity $\hat{\vect z}\times\hat{\vect r}=\hat{\vect\varphi}$ , which is the right handed triad drawn in the second panel above; since $\vect d$ and $\hat{\vect r}$ are both horizontal, $\vect d\times\hat{\vect r}$ is purely vertical, so the two terms point along the mutually perpendicular directions $\hat{\vect z}$ and $\hat{\vect\varphi}$ and can only cancel each other by both vanishing.

This torque is zero at every instant only when $\vect d=\vect 0$ and $z_P=0$ , that is, only at $P=C$ .

\vect\tau_P = 0\ \text{for all }t \iff P = C

(2)

Since $O$ sits a height $l\cos\theta\neq0$ above $C$ , $\vect M_O$ is not constant; only $\vect M_C$ , the angular momentum about the circle’s own center, is.

Only the circle’s own center

C

; nowhere else,

O

included.

Step 3Angular momentum about

O

, split into a fixed and a rotating part

Writing the ball’s position relative to $O$ as $r\hat{\vect r}-l\cos\theta\,\hat{\vect z}$ (measuring from $C$ ’s height) and its velocity as $r\omega\,\hat{\vect\varphi}$ ,

\begin{align*}\vect M_O &= m\left(r\hat{\vect r}-l\cos\theta\,\hat{\vect z}\right)\times\left(r\omega\,\hat{\vect\varphi}\right) \\ &= mr^2\omega\,(\hat{\vect r}\times\hat{\vect\varphi}) - ml\omega r\cos\theta\,(\hat{\vect z}\times\hat{\vect\varphi}) \\ &= mr^2\omega\,\hat{\vect z} + ml\omega r\cos\theta\,\hat{\vect r},\end{align*}

using $\hat{\vect r}\times\hat{\vect\varphi}=\hat{\vect z}$ and $\hat{\vect z}\times\hat{\vect\varphi}=-\hat{\vect r}$ . The vertical term is exactly $\vect M_C$ and stays fixed; the second term has constant magnitude $ml\omega r\cos\theta$ but rotates with the ball, always pointing radially outward along $\hat{\vect r}(t)$ .

Step 4The change over half a revolution

After half a revolution $\hat{\vect r}$ reverses, $\hat{\vect r}(t_0+T/2)=-\hat{\vect r}(t_0)$ , while the vertical part is unchanged, so only the rotating part contributes to the change.

\Delta\vect M_O = ml\omega r\cos\theta\left[\hat{\vect r}(t_0+T/2)-\hat{\vect r}(t_0)\right] = -2ml\omega r\cos\theta\,\hat{\vect r}(t_0).

Taking the magnitude and substituting $r=l\sin\theta$ ,

|\Delta\vect M_O| = 2ml\omega r\cos\theta = 2m\omega l^2\sin\theta\cos\theta.

Substituting $\cos\theta=g/(\omega^2 l)$ from (1) and $\sin\theta=\sqrt{1-\cos^2\theta}$ into this,

\begin{align*}|\Delta\vect M_O| &= 2m\omega l^2\cdot\sqrt{1-\left(\dfrac{g}{\omega^2 l}\right)^2}\cdot\dfrac{g}{\omega^2 l}\end{align*}

|\Delta\vect M_O| = \dfrac{2mgl}{\omega}\sqrt{1-\left(\dfrac{g}{\omega^2 l}\right)^2}

(3)

Check

Dimensionally $mgl/\omega$ is (kg)(m/s $^2$ )(m)/(rad/s) = kg m $^2$ /s, an angular momentum, and the square root is dimensionless as required. As $\omega\to\infty$ the cone flattens ( $\cos\theta\to0$ , $\sin\theta\to1$ ) and the bracket approaches 1 while the prefactor $2mgl/\omega\to0$ , so $|\Delta\vect M_O|\to0$ : an infinitely fast, infinitely flat circle sweeps through no vertical extent for gravity’s torque to act on.

Remark

The rotating part of $\vect M_O$ traces its own circle of radius $ml\omega r\cos\theta$ as the ball goes around, so integrated over one full revolution it returns to its starting value and $\Delta\vect M_O=0$ for a whole period; only a fractional turn, such as this half revolution, picks up a nonzero net change, and the same half-turn calculation reproduces the answer for any starting instant $t_0$ since $\hat{\vect r}(t_0)$ cancels out of the magnitude.