Irodov 1.30: Two Accelerations Trade Places Mid Flight

Problem Irodov 1.30

For the same projectile as before, launched at speed $v_0$ and angle $\alpha$ above the horizontal under constant gravity $g$ with no air drag, sketch how three quantities vary over the flight, from launch ( $t=0$ ) to landing ( $t=\tau$ ): the modulus of the normal acceleration $w_n$ , the modulus of the tangential acceleration $w_\tau$ , and the signed projection $w_v$ of the total acceleration onto the velocity direction.

Given

$v_0$
$\alpha$
$g$ (as in the foregoing problem)

Find

the shapes of

w_n(t)

w_\tau(t)

w_v(t)

over

0\le t\le\tau

Solution

Step 1Velocity components through the flight

Only gravity acts, so the horizontal component of velocity is constant while the vertical component decreases linearly; both come from one integration of $\vect g$ .

v_x = v_0\cos\alpha, \qquad v_y(t) = v_0\sin\alpha-gt, \qquad v(t)=\sqrt{v_x^2+v_y(t)^2} \tag{1}

Step 2The signed projection of

\vect g

on the velocity

Projecting the constant acceleration $\vect g=-g\vect j$ onto the unit vector $\vect v/v$ isolates this component directly.

\begin{align*}w_v &= \vect g\cdot\frac{\vect v}{v} = \frac{-g\,v_y(t)}{v(t)}\end{align*}

w_v(t) = \dfrac{g\bigl(gt-v_0\sin\alpha\bigr)}{\sqrt{v_0^2\cos^2\alpha+(v_0\sin\alpha-gt)^2}}

(2)

Step 3Tangential and normal moduli from the same projection

The tangential modulus is the same component read as a magnitude, so (2) already carries it:

w_\tau(t) = |w_v(t)|

The normal modulus is the component of $\vect g$ perpendicular to $\vect v$ . Taking the cross product of the two vectors and dividing by $v$ isolates that component, and with $\vect g = -g\vect j$ only the horizontal piece of $\vect v$ survives the product.

\begin{align*}w_n &= \frac{\bigl|\vect g\times\vect v\bigr|}{v} = \frac{\bigl|(-g\vect j)\times(v_x\vect i+v_y\vect j)\bigr|}{v(t)}\\ &= \frac{g\,v_x}{v(t)}\end{align*}

Substituting $v_x=v_0\cos\alpha$ and $v(t)$ from (1) gives the normal modulus in terms of the givens.

w_n(t) = \dfrac{g\,v_0\cos\alpha}{\sqrt{v_0^2\cos^2\alpha+(v_0\sin\alpha-gt)^2}}

(3)

Squaring and adding (2) and (3) shows the pair accounts for the whole of $\vect g$ , since $v_x^2+v_y^2=v^2$ :

w_n^2+w_v^2 = \frac{g^2 v_x^2}{v^2}+\frac{g^2 v_y^2}{v^2} = g^2 \tag{4}

Step 4Values at the launch, the peak and the landing

The three curves over one flight, sketched for α=30°: w_n (gold), w_τ (royal) and w_v (sky). w_τ is dashed from the peak onward, where w_v crosses zero and the two run together. — The three curves over one flight, sketched for $\alpha=30^\circ$ : $w_n$ (gold), $w_\tau$ (royal) and $w_v$ (sky). $w_\tau$ is dashed from the peak onward, where $w_v$ crosses zero and the two run together.

At $t=0$ and $t=\tau$ , $v(t)=v_0$ by the symmetry of (1); at the peak $t=\tau/2$ , $v_y=0$ so $v(t)=v_0\cos\alpha$ . Putting those three speeds into (3) and (2), and taking $w_\tau=|w_v|$ ,

\begin{align*}w_n(0)=w_n(\tau) &= g\cos\alpha, \qquad w_n(\tau/2) = g \\ w_\tau(0)=w_\tau(\tau) &= g\sin\alpha, \qquad w_\tau(\tau/2) = 0 \\ w_v(0) &= -g\sin\alpha,\qquad w_v(\tau/2)=0, \qquad w_v(\tau) = g\sin\alpha\end{align*}

Over the flight, $w_n$ rises smoothly from $g\cos\alpha$ to $g$ at the peak and back to $g\cos\alpha$ at landing; $w_\tau$ falls from $g\sin\alpha$ to $0$ at the peak and climbs back, a V touching the axis; and $w_v$ runs from $-g\sin\alpha$ through $0$ at the peak up to $+g\sin\alpha$ , crossing zero exactly where $w_\tau$ bottoms out, all sketched in the figure.

Three curves sketched in the figure above

Check

At every instant $w_n^2+w_v^2=g^2$ by (4), so the total acceleration always has magnitude $g$ , as it must for free fall; at the peak this gives $w_n=g$ , $w_v=0$ exactly, matching Step 4.

Remark

$w_\tau$ and $w_v$ carry the same information, $|w_v|$ against $w_v$ , but the modulus throws away the one fact worth keeping: whether the body is still speeding up or already slowing down. A speedometer reads the speed $v$ itself, never $w_\tau$ directly; it is the sign of $w_v$ , not $w_\tau$ , that tells you which side of the peak the body is on.