Irodov 1.189: Angular Momentum Seen from a Sliding Frame

Problem Irodov 1.189

A ball of mass $m$ is released from rest and falls a height $h$ under gravity, air drag neglected. A second reference frame carries a fixed point $O$ that translates horizontally at constant velocity $\vect V$ ; the ball is released exactly from $O$ ’s instantaneous position at the moment of release. Find the magnitude of the change in the ball’s angular momentum, measured relative to the moving point $O$ and using the ball’s velocity relative to that frame, accumulated over the whole time it takes the ball to fall the height $h$ .

At time t the ball has fallen 12 gt^2 while the frame's point O has moved a horizontal distance Vt from the common release point. The vector r ' joins O to the ball: its position relative to the moving frame. — At time $t$ the ball has fallen $\frac12 gt^2$ while the frame’s point $O$ has moved a horizontal distance $Vt$ from the common release point. The vector $\vect r\,'$ joins $O$ to the ball: its position relative to the moving frame.

Given

$m$
$h$
$\vect V$
$g$

Find

|\Delta\vect M_O|

over the fall

Solution

Step 1Position and velocity relative to the moving frame

In the ground frame the ball falls from rest, so at time $t$ its velocity is $gt$ downward and its displacement is $gt^2/2$ downward, while $O$ has moved a horizontal distance $Vt$ from the common start. Subtracting the frame’s own uniform translation gives the ball’s position and velocity relative to $O$ ,

\begin{align*}\vect r\,'(t) &= -Vt\,\vect i + \frac{1}{2}gt^2\,\vect j \\ \vect v\,'(t) &= -V\,\vect i + gt\,\vect j,\end{align*}

with $\vect i$ horizontal (the direction of $\vect V$ ) and $\vect j$ vertically down.

Step 2Angular momentum relative to

O

as a function of time

The angular momentum relative to $O$ , using the relative position and the relative velocity, is $\vect M_O(t) = m\,\vect r\,'(t)\times\vect v\,'(t)$ .

Only the mixed terms remain, and both of them lie along $\vect i\times\vect j=\vect k$ ,

\begin{align*}\vect M_O(t) &= m\left(-Vt\,\vect i+\frac{1}{2}gt^2\,\vect j\right)\times\left(-V\,\vect i+gt\,\vect j\right) \\ &= m\left[-Vt\cdot gt\,(\vect i\times\vect j) + \frac{1}{2}gt^2\cdot(-V)\,(\vect j\times\vect i)\right] \\ &= m\left[-Vgt^2 + \frac{1}{2}Vgt^2\right]\vect k\end{align*}

\vect M_O(t) = -\frac{1}{2}mVgt^2\,\vect k

(1)

Since $\vect M_O(0)=\vect 0$ (the ball starts exactly at $O$ ), this expression is itself the increment picked up by time $t$ .

Step 3Evaluate at the landing time

The fall from rest through height $h$ takes $t_f=\sqrt{2h/g}$ , found from $h=gt_f^2/2$ .

t_f^2 = \frac{2h}{g}

Substituting into (1) and taking the magnitude,

\begin{align*}|\vect M_O(t_f)| &= \frac{1}{2}mVg\,t_f^2 \\ &= \frac{1}{2}mVg\cdot\frac{2h}{g}\end{align*}

|\Delta\vect M_O| = mVh

(2)

Check

Dimensionally $[m][V][h] = \text{kg}\cdot(\text{m/s})\cdot\text{m} = \text{kg}\,\text{m}^2/\text{s}$ , an angular momentum, as required. Also, if $V=0$ the moving frame coincides with the ground frame and the ball falls straight through $O$ with $\vect r\,'\parallel\vect v\,'$ at every instant, giving $\vect M_O\equiv\vect 0$ ; the boxed result correctly vanishes at $V=0$ .

Remark

The formula (1) for $\vect M_O(t)$ shows the angular momentum growing quadratically with time along the way, at a rate set entirely by how far the frame has drifted while gravity does the same free fall it always does; only at the specific instant of landing does it collapse to the clean form $mVh$ . The one-half coefficient comes from a partial cancellation inside the cross product: pairing the horizontal leg of $\vect r\,'$ with the vertical leg of $\vect v\,'$ contributes $-mVgt^2\,\vect k$ , while pairing the vertical leg of $\vect r\,'$ with the horizontal leg of $\vect v\,'$ contributes $+mVgt^2/2\,\vect k$ in the opposite sense; adding the two leaves the net $-mVgt^2/2\,\vect k$ of (1).