Irodov 1.125: Power Oscillates Yet Averages to Zero – Amit & Deepali Sinha Foundation

Problem Irodov 1.125

A body of mass $m$ is launched with initial speed $v_0$ at an angle $\alpha$ above the horizontal and moves freely under constant gravitational acceleration $g$ , air resistance neglected, until it returns to its launch height. Find the instantaneous power $P(t)$ delivered by gravity as a function of the elapsed time $t$ , and the mean power $\langle P\rangle$ developed by gravity averaged over the whole time of motion, from launch to landing.

The parabolic flight path, launched with v_0 at the angle α. At a general point A on the way up, the velocity v and the weight m g are both drawn; gravity's power there is their dot product. — The parabolic flight path, launched with $\vect v_0$ at the angle $\alpha$ . At a general point $A$ on the way up, the velocity $\vect v$ and the weight $m\vect g$ are both drawn; gravity’s power there is their dot product.

Given

$m$
$v_0$
$\alpha$

Find

P(t)

\langle P\rangle

over the whole flight

Solution

Step 1Vertical velocity component

Taking upward as positive, gravity acts alone on the vertical component of motion, so integrating the constant acceleration $-g$ from the launch value $v_0\sin\alpha$ gives $v_y(t)$ .

v_y(t) = v_0\sin\alpha-\int_0^t g\,d\tau

v_y(t) = v_0\sin\alpha-gt

(1)

Step 2Instantaneous power of gravity

Gravity’s force is $-mg$ along the vertical, so its power is the product of that force with the vertical velocity component alone, the horizontal component contributing nothing to the dot product.

P(t) = (-mg)\,v_y(t) = -mg\bigl(v_0\sin\alpha-gt\bigr)

P(t) = mg(gt-v_0\sin\alpha)

(2)

Step 3Time of flight

The body lands when it returns to launch height, so the vertical displacement $y(t)=v_0\sin\alpha\,t-gt^2/2$ returns to zero at the nonzero root $T$ .

0 = v_0\sin\alpha\,T-\frac{gT^2}{2}

T = \dfrac{2v_0\sin\alpha}{g}

(3)

Step 4Mean power over the whole flight

The mean power is the total work of gravity divided by the flight time (3); since the body lands at the same height it was launched from, gravity’s net work over the whole flight is zero.

\langle P\rangle = \frac{1}{T}\int_0^T P(t)\,dt = \frac{A_{\mathrm{grav}}}{T} = \frac{-mg\cdot0}{T}

\langle P\rangle = 0

(4)

Check

Dimensionally $[m][g][g][t]=\mathrm{kg}\cdot(\mathrm{m/s^2})^2\cdot\mathrm s=\mathrm{kg\,m^2/s^3}=\mathrm W$ , a power, as required. At $t=0$ , $P(0)=-mgv_0\sin\alpha<0$ for any upward launch, correctly negative since gravity opposes the initial upward climb; $P$ then rises linearly and crosses zero exactly at $t=T/2$ , the instant of maximum height, where $v_y=0$ by (1) and gravity is momentarily doing no work at all.

Remark

$P(t)$ is linear in $t$ while the flight is symmetric about $t=T/2$ in every other respect, so the negative power on the way up and the positive power on the way down cancel exactly on averaging, independent of $\alpha$ or $v_0$ ; only the endpoints of the flight, both at the launch height, matter for $\langle P\rangle$ , never the shape of the path between them. The same zero mean would hold for any conservative force with a time-independent potential energy, not just gravity, whenever the body returns to a point of equal potential energy: mean power is always total work over elapsed time, and here that total work is forced to vanish by the landing condition alone.