Irodov 1.245: Work Beats a Varying Torque – Amit & Deepali Sinha Foundation

Problem Irodov 1.245

A thin uniform rod $AB$ of mass $m$ and length $l$ lies in a horizontal plane and can rotate freely about a vertical axis through its end $A$ . Starting from rest, the end $B$ is subjected to a constant horizontal force $F$ whose direction stays fixed, perpendicular to the rod’s original (starting) position, throughout the motion. Find the rod’s angular velocity $\omega$ as a function of the angle $\varphi$ it has turned through, measured from its initial position.

Seen from above. The rod has turned through φ from its original position, drawn dashed; F keeps the direction it had at the start, perpendicular to that dashed line, and stays applied at the end B. — Seen from above. The rod has turned through $\varphi$ from its original position, drawn dashed; $\vect F$ keeps the direction it had at the start, perpendicular to that dashed line, and stays applied at the end $B$ .

Given

$m$
$l$
$F$ constant in magnitude and direction
$\omega = 0$ at $\varphi = 0$

Find

Find

\omega(\varphi)

Solution

Step 1Why energy and not torque

The rod turns about a fixed vertical axis through $A$ , and gravity, the axis reaction and the normal support all do no work: the weight is vertical while every displacement is horizontal, and the reaction acts at the point that does not move. The only agent doing work is $\vect F$ . So the work $\vect F$ has done by the time the rod has turned through $\varphi$ equals the rod’s kinetic energy there, and that is a direct route to $\omega(\varphi)$ .

Step 2The work done in turning through

d\varphi

Set up coordinates in the horizontal plane with the origin at $A$ and the $x$ axis along the rod’s original position, so that $\vect F = F\,\vect{\hat y}$ throughout. When the rod has turned through $\varphi$ , the end $B$ sits at

\vect r_B = l\,(\cos\varphi,\ \sin\varphi),

and a further turn $d\varphi$ moves it by

d\vect r_B = \frac{d\vect r_B}{d\varphi}\,d\varphi = l\,(-\sin\varphi,\ \cos\varphi)\,d\varphi .

The elementary work is the scalar product of the constant force with this displacement,

dA = \vect F \cdot d\vect r_B = Fl\cos\varphi\,d\varphi .

The same instant, with the moment arm of F about the axis drawn as the perpendicular from A onto the force's line of action. Its length is lφ, which is exactly the coefficient appearing in dA. — The same instant, with the moment arm of $\vect F$ about the axis drawn as the perpendicular from $A$ onto the force’s line of action. Its length is $l\cos\varphi$ , which is exactly the coefficient appearing in $dA$ .

Step 3Total work up to the current angle

Integrate from the initial position to the current one:

A = \int_{0}^{\varphi} Fl\cos\varphi'\,d\varphi' = Fl\bigl[\sin\varphi'\bigr]_{0}^{\varphi} .

A = Fl\sin\varphi

(1)

Step 4Convert the work into spin

The rod starts from rest, so all of $A$ has gone into rotational kinetic energy about the axis through the end $A$ . For a thin uniform rod about a perpendicular axis through one end,

I = \frac{ml^{2}}{3},

and the energy balance $\frac{1}{2}I\omega^{2} = A$ reads

\frac{1}{2}\cdot\frac{ml^{2}}{3}\,\omega^{2} = Fl\sin\varphi .

Solving for $\omega$ , the factor $l$ cancels once:

\omega = \sqrt{\frac{6F\sin\varphi}{ml}}

(2)

Check

Test the start of the motion, where the answer can be got another way. Near $\varphi = 0$ the moment arm is the full $l$ , so the angular acceleration is $\beta = Fl/I = 3F/(ml)$ and constant to first order; uniformly accelerated rotation from rest gives $\omega^{2} = 2\beta\varphi = 6F\varphi/(ml)$ , which is the answer with $\sin\varphi$ replaced by $\varphi$ , as it should be for small $\varphi$ . The dimensions also work: $F/(ml)$ has units of $\mathrm{s^{-2}}$ .

Remark

The formula stops at $\varphi = \pi$ , where $\omega$ returns to zero. Past $\varphi = \pi/2$ the moment arm $l\cos\varphi$ is negative, the force starts taking energy back, and at half a turn the rod is momentarily at rest again and then swings back. The motion is an oscillation about $\varphi = \pi/2$ , not a spin up.