Irodov 1.186: Angular Momentum Grows as Time Squared

Problem Irodov 1.186

A ball of mass $m$ is launched from a point $O$ with initial speed $v_0$ at angle $\alpha$ above the horizontal, moving freely under constant gravitational acceleration $g$ with air drag neglected. Find the magnitude of the ball’s angular momentum about $O$ as a function of the elapsed time $t$ . Then evaluate this angular momentum at the instant the ball reaches the highest point of its trajectory, for $m=130\text{ g}$ , $\alpha=45^\circ$ , $v_0=25\text{ m/s}$ .

The parabolic flight path from the launch point O, showing the position r and velocity v at the general point P reached at time t. At P_max, the highest point, v is horizontal. — The parabolic flight path from the launch point $O$ , showing the position $\vect r$ and velocity $\vect v$ at the general point $P$ reached at time $t$ . At $P_{\mathrm{max}}$ , the highest point, $\vect v$ is horizontal.

Given

$m$
$\alpha$
$\vect v_0$
$g$

Find

M(t)

;

M

at the highest point for

m=130\text{ g}

\alpha=45^\circ

v_0=25\text{ m/s}

Solution

Step 1Position and velocity from launch

Take $O$ as the origin with $x$ horizontal and $y$ vertically upward, and let $\vect i$ , $\vect j$ , $\vect k$ be the unit vectors along $x$ , $y$ and the outward normal to the plane of flight. The standard projectile equations under constant $g$ give the four components.

\begin{aligned} x(t) &= v_0\cos\alpha\, t, & y(t) &= v_0\sin\alpha\, t - \frac{1}{2}gt^2, \\ v_x(t) &= v_0\cos\alpha, & v_y(t) &= v_0\sin\alpha - gt. \end{aligned} \tag{1}

The two vectors the angular momentum is built from are those components on the two in-plane unit vectors, with no $\vect k$ part at any time, since the flight never leaves the plane.

\begin{aligned} \vect r(t) &= x(t)\,\vect i + y(t)\,\vect j \\ \vect v(t) &= v_x(t)\,\vect i + v_y(t)\,\vect j \end{aligned} \tag{2}

Step 2The angular momentum as a cross product

The angular momentum about $O$ is $\vect M = m\,\vect r\times\vect v$ . Expanding that cross product from (2) gives four terms, one for each pairing of the two unit vectors.

\begin{align*} \vect r\times\vect v &= (x\,\vect i + y\,\vect j)\times(v_x\,\vect i + v_y\,\vect j) \\ &= xv_x(\vect i\times\vect i) + xv_y(\vect i\times\vect j) + yv_x(\vect j\times\vect i) + yv_y(\vect j\times\vect j) \\ &= (xv_y - yv_x)\,\vect k \tag{3}\end{align*}

Substituting the components of (1) into (3) and multiplying out both products,

\begin{align*}xv_y &= v_0\cos\alpha\,t\,(v_0\sin\alpha-gt) = v_0^2\sin\alpha\cos\alpha\,t - gv_0\cos\alpha\,t^2, \\ yv_x &= \left(v_0\sin\alpha\,t-\frac{1}{2}gt^2\right)v_0\cos\alpha = v_0^2\sin\alpha\cos\alpha\,t - \frac{1}{2}gv_0\cos\alpha\,t^2.\end{align*}

Subtracting cancels the identical first terms, and (3) turns the difference into the one surviving component of $\vect M$ .

\begin{aligned} xv_y - yv_x &= -gv_0\cos\alpha\,t^2 + \frac{1}{2}gv_0\cos\alpha\,t^2 = -\frac{1}{2}gv_0\cos\alpha\,t^2 \\ M_z(t) = m(xv_y-yv_x) &= -\frac{1}{2}mgv_0\cos\alpha\,t^2 \end{aligned} \tag{4}

Only the magnitude of the angular momentum is asked for, so

M(t) = |M_z(t)| = \dfrac{1}{2}mgv_0t^2\cos\alpha

(5)

Step 3The time of flight to the highest point

The highest point is reached when the vertical velocity vanishes, $v_y=0$ in (1), giving

t_{\mathrm{top}} = \frac{v_0\sin\alpha}{g}.

Step 4The angular momentum at the highest point

Substituting $t_{\mathrm{top}}$ into (5) gives $M$ at the apex.

\begin{align*}M_{\mathrm{top}} = \frac{1}{2}mgv_0\cos\alpha\left(\frac{v_0\sin\alpha}{g}\right)^2 &= \frac{mv_0^3\sin^2\alpha\cos\alpha}{2g}\end{align*}

With $m=0.130\text{ kg}$ , $\alpha=45^\circ$ ( $\sin^2\alpha=1/2$ , $\cos\alpha=\sqrt2/2$ ), $v_0=25\text{ m/s}$ and $g=9.8\text{ m/s}^2$ substituted in,

M_{\mathrm{top}} = \frac{(0.130)(25)^3(1/2)(\sqrt2/2)}{2(9.8)}\ \text{kg}\cdot\text{m}^2/\text{s}

M_{\mathrm{top}} = \dfrac{mv_0^3\sin^2\alpha\cos\alpha}{2g} = 37\text{ kg}\cdot\text{m}^2/\text{s}

(6)

Check

Differentiating the signed result (4) gives $dM_z/dt=-mgv_0\cos\alpha\,t$ . Independently, gravity is the only force present, and a force $-mg\hat{\vect j}$ acting at horizontal offset $x(t)$ exerts torque $N=-mgx(t)=-mgv_0\cos\alpha\,t$ about $O$ : the two expressions for $dM_z/dt$ agree, confirming Step 2 without repeating its cross-product algebra. Also $M(0)=0$ , as it must be, since $\vect r=\vect 0$ at the launch point itself.

Remark

The torque of gravity about $O$ grows only linearly with time, $N=-mgv_0\cos\alpha\,t$ , yet the angular momentum it builds grows quadratically, $M\propto t^2$ , because $M$ is the running time-integral of that torque. The same pattern appears in $x(t)$ of (1), which is linear in $t$ while $y(t)$ (built from the same constant $g$ ) carries a $t^2$ term: one more power of $t$ appears every time a constant rate is integrated once more.