Irodov 1.69: Work Only Sees the Angle – Amit & Deepali Sinha Foundation

Problem Irodov 1.69

A bar of mass $m$ lies on a smooth (frictionless) horizontal plane. Starting from rest, it is pulled by a force of fixed magnitude $F = mg/3$ whose direction stays in the vertical plane of the bar’s straight-line motion but steadily tilts: the angle $\alpha$ between the force and the horizontal grows with the distance $s$ already covered, according to $\alpha = as$ for a constant $a$ . Find the bar’s speed $v$ as a function of $\alpha$ .

The bar mid-slide on the smooth plane: the applied force F of fixed magnitude tilts at angle α=as above the horizontal, set by the distance s the bar has already advanced along its line of motion. — The bar mid-slide on the smooth plane: the applied force $\vect F$ of fixed magnitude tilts at angle $\alpha=as$ above the horizontal, set by the distance $s$ the bar has already advanced along its line of motion.

Given

$F = mg/3$
$\alpha = as$
$m$
no friction

Find

v(\alpha)

Solution

Step 1Only the along-track component of

\vect F

does work

The three forces on the bar: the applied F, the weight m g, and the plane's normal reaction N. Only F has a component along the motion. — The three forces on the bar: the applied $\vect F$ , the weight $m\vect g$ , and the plane’s normal reaction $\vect N$ . Only $\vect F$ has a component along the motion.

The bar’s motion is confined to the horizontal line it started on, so only the component of $\vect F$ along that line, $F\cos\alpha$ , does work; the perpendicular component and the (frictionless) normal reaction do none. The work-energy theorem over an element $ds$ of the path reads

d\!\left(\dfrac{mv^2}{2}\right) = F\cos\alpha\,ds.

Step 2Trade the path length for the angle

Since $\alpha=as$ , the distance and the angle are proportional, $ds = d\alpha/a$ , so the work integral becomes an integral over $\alpha$ alone, run from the start ( $\alpha=0$ ) up to the angle asked for.

\int_0^{v} mv'\,dv' = \int_0^{\alpha} F\cos\alpha'\,\frac{d\alpha'}{a} \tag{1}

\begin{align*}\frac{mv^2}{2} &= \frac{F}{a}\int_0^{\alpha}\cos\alpha'\,d\alpha' \\ &= \frac{F}{a}\Bigl[\sin\alpha'\Bigr]_0^{\alpha}\end{align*}

\dfrac{mv^2}{2} = \dfrac{F}{a}\sin\alpha

(2)

Step 3Insert the given force and solve for

v

Substituting $F=mg/3$ into (2) and solving for $v$ ,

\begin{align*}\frac{mv^2}{2} &= \frac{mg}{3a}\sin\alpha\\ v^2 &= \frac{2g}{3a}\sin\alpha\end{align*}

v = \sqrt{\dfrac{2g}{3a}\sin\alpha}

(3)

Check

Dimensionally, $g$ has units of $\mathrm{m/s^2}$ and $\alpha=as$ is dimensionless, so $a$ has units of $\mathrm{m}^{-1}$ and $g/a$ has units of $\mathrm{m^2/s^2}$ , a speed squared, matching $v^2$ . For small $\alpha$ , $\sin\alpha\approx\alpha=as$ , and the boxed result reduces to $v^2\approx 2gs/3=2(F/m)s$ , exactly the constant-force result for a force held at the fixed angle $\alpha\approx0$ , as it must for a bar that has moved only a short distance.

Remark

$\alpha=90^\circ$ is not a boundary of validity, only the point where $v$ peaks: $N=mg-(mg/3)\sin\alpha$ never drops below $2mg/3$ , so the bar never leaves the plane at any $\alpha$ , and the work integral (1) is exact for the whole range $0\le\alpha\le180^\circ$ . Past $90^\circ$ the horizontal component $F\cos\alpha$ turns negative and decelerates the bar, but the bar keeps advancing until $v$ itself reaches zero, exactly at $\alpha=180^\circ$ ( $s=\pi/a$ ), where $\sin\alpha$ returns to $0$ . On that far side the same speed $v$ is reached at two different angles, $\alpha$ and $180^\circ-\alpha$ , one while $F$ is still speeding the bar up and one while it is slowing it down, so $\alpha$ alone does not say which branch of the motion the bar is on.