Irodov 1.97: The Force That Rises Then Fades – Amit & Deepali Sinha Foundation

Problem Irodov 1.97

A particle of mass $m$ is at rest at $t=0$ . From that instant until $t=\tau$ it experiences a force $\vect F = \vect a\,t(\tau - t)$ , where $\vect a$ is a constant vector and $\tau$ is the fixed duration over which the force acts; after $t=\tau$ the force is gone. Find:

the particle’s momentum $\vect p$ at the instant $t=\tau$ , when the force stops acting
the distance $s$ the particle covers during the interval $0\le t\le\tau$ while the force acts

The force's magnitude against time, in units where a=1, τ=1: F=t(τ-t) vanishes at both t=0 and t=τ and peaks at the midpoint t=τ/2, so the pulse switches on and off smoothly with no jump in F at either end. — The force’s magnitude against time, in units where $a=1$ , $\tau=1$ : $F=t(\tau-t)$ vanishes at both $t=0$ and $t=\tau$ and peaks at the midpoint $t=\tau/2$ , so the pulse switches on and off smoothly with no jump in $F$ at either end.

Given

$m$
$\vect a$
$\tau$
particle at rest at $t=0$

Find

(a)

\vect p

t=\tau

, (b)

s

covered over

0\le t\le\tau

Solution

Step 1Momentum from the impulse of the force

Newton’s second law in the form $d\vect p/dt=\vect F$ integrates directly to the momentum, since the particle starts from rest. The force never changes direction (it is $\vect a$ times the scalar $t(\tau-t)\ge0$ on $[0,\tau]$ ), so the motion stays along $\vect a$ throughout.

\begin{align*}\vect p(\tau) &= \int_0^\tau \vect a\,t(\tau-t)\,dt \\ &= \vect a\int_0^\tau \bigl(\tau t - t^2\bigr)\,dt \\ &= \vect a\left.\left(\frac{\tau t^2}{2}-\frac{t^3}{3}\right)\right|_0^\tau \\ &= \vect a\left(\frac{\tau^3}{2}-\frac{\tau^3}{3}\right)\end{align*}

\vect p = \dfrac{\vect a\tau^3}{6}

(1)

Step 2Velocity as a function of time

Dividing the general momentum $\vect p(t)=\vect a(\tau t^2/2-t^3/3)$ from Step 1 by $m$ gives the velocity at any time in $[0,\tau]$ :

\vect v(t) = \frac{\vect a}{m}\left(\frac{\tau t^2}{2}-\frac{t^3}{3}\right). \tag{2}

Step 3Distance covered while the force acts

The distance is the integral of speed over the interval the force acts on, $[0,\tau]$ . The scalar factor in (2) never changes sign there,

\frac{\tau t^2}{2}-\frac{t^3}{3} = \frac{t^2(3\tau-2t)}{6} \ge 0 \qquad (0\le t\le\tau),

so $\vect v(t)$ never reverses direction on this interval and the path length equals the magnitude of the net displacement.

\begin{align*}s &= \int_0^\tau v(t)\,dt \\ &= \frac{a}{m}\int_0^\tau\left(\frac{\tau t^2}{2}-\frac{t^3}{3}\right)dt \\ &= \frac{a}{m}\left.\left(\frac{\tau t^3}{6}-\frac{t^4}{12}\right)\right|_0^\tau \\ &= \frac{a}{m}\left(\frac{\tau^4}{6}-\frac{\tau^4}{12}\right)\end{align*}

s = \dfrac{a\tau^4}{12m}

(3)

Check

Dimensionally, $[F]=[a][t]^2$ forces $[a]=\mathrm{N/s^2}$ , so $[a][\tau]^3$ is $\mathrm{N\cdot s}$ , a momentum, and $[a][\tau]^4/[m]$ is $\mathrm{N\cdot s^2/kg}=\mathrm m$ , a length; both boxed results check. Also, differentiating $\vect p(t)=\vect a(\tau t^2/2-t^3/3)$ gives back $d\vect p/dt=\vect a(\tau t-t^2)=\vect a\,t(\tau-t)$ , exactly the applied force, confirming Step 1’s integration.

Remark

The impulse $\vect a\tau^3/6$ is exactly two thirds of $\vect F_{\mathrm{max}}\tau$ , the peak force times the full duration: $F_{\mathrm{max}}=a(\tau/2)(\tau-\tau/2)=a\tau^2/4$ at the midpoint (marked on the plot above), and $(a\tau^2/4)\cdot\tau\cdot 2/3=a\tau^3/6$ reproduces the impulse exactly. That $2/3$ factor is the classical area of a parabolic segment against its bounding rectangle, here appearing as the fraction of the peak-force-times-duration product that a parabolic pulse actually delivers.