Irodov 1.234: No Rotation Locates the Mass Center – Amit & Deepali Sinha Foundation

Problem Irodov 1.234

A thin uniform rod $AB$ of mass $m = 1.0$ kg moves translationally, that is, without turning at all, with acceleration $w = 2.0$ m/s $^2$ . Two antiparallel forces perpendicular to the rod drive it: $\vect F_1$ is applied at the end $A$ , and $\vect F_2$ is applied at a point of the rod a distance $a = 20$ cm from $A$ in the direction of $B$ , the second force having magnitude $F_2 = 5.0$ N. Find the length of the rod.

The rod m, end A at the top, B at the bottom. F_1 acts at A and F_2 a distance a farther along, pointing the opposite way across the rod. The acceleration w is drawn clear of the body. — The rod $m$ , end $A$ at the top, $B$ at the bottom. $\vect F_1$ acts at $A$ and $\vect F_2$ a distance $a$ farther along, pointing the opposite way across the rod. The acceleration $\vect w$ is drawn clear of the body.

Given

$m = 1.0$ kg
$w = 2.0$ m/s $^2$
$a = 20$ cm $= 0.20$ m
$F_2 = 5.0$ N

Find

l

, the length of the rod

Solution

Step 1The word “translationally” is two equations

A body that translates keeps its orientation fixed, so its angular velocity and its angular acceleration are both zero at every instant. Its equation of motion therefore splits into two independent statements: Newton’s second law for the center of mass, and the vanishing of the net torque about that center. Along the common line of the two forces the first reads

F_2 - F_1 = mw ,

the difference of the magnitudes because the pair is antiparallel.

F_1 = F_2 - mw

(1)

Step 2The moment arms about the center of mass

The second statement needs the two arms. A uniform rod has its center of mass at its midpoint $C$ , both forces are perpendicular to the rod, and both act on the same side of $C$ , so each arm is simply the distance from $C$ to the point of application, measured along the rod.

The moment arms about the center of mass C, the midpoint of the uniform rod: l/2 from C up to the end A where F_1 acts, and l/2-a from C up to the point where F_2 acts. — The moment arms about the center of mass $C$ , the midpoint of the uniform rod: $l/2$ from $C$ up to the end $A$ where $\vect F_1$ acts, and $l/2-a$ from $C$ up to the point where $\vect F_2$ acts.

The two torques about $C$ have opposite senses, and translation demands that they cancel.

F_1\cdot\frac{l}{2} - F_2\left(\frac{l}{2}-a\right) = 0

F_1\,\frac{l}{2} = F_2\left(\frac{l}{2}-a\right)

(2)

The arm of $\vect F_1$ is the longer one, so this balance already forces $F_1<F_2$ , and with it the positive $mw$ in (1): the rod accelerates along $\vect F_2$ .

Step 3Eliminate the unknown force and solve for the length

Only $l$ is wanted, so $F_1$ goes: substituting (1) into (2) leaves one equation in $l$ , and the $F_2l/2$ terms cancel from the two sides.

\begin{align*}\left(F_2-mw\right)\frac{l}{2} &= F_2\left(\frac{l}{2}-a\right) \\ F_2\,\frac{l}{2}-mw\,\frac{l}{2} &= F_2\,\frac{l}{2}-F_2a \\ mw\,\frac{l}{2} &= F_2 a\end{align*}

l = \frac{2aF_2}{mw} = \frac{2(0.20\ \mathrm{m})(5.0\ \mathrm{N})}{(1.0\ \mathrm{kg})(2.0\ \mathrm{m/s^2})} = 1.0\ \mathrm{m}

(3)

Check

Dimensionally $aF_2/(mw)$ is a length, since $mw$ and $F_2$ are both forces. Numerically the two conditions are consistent: (1) gives $F_1 = 5.0-2.0 = 3.0$ N, and with $l = 1.0$ m the arms are $0.50$ m and $0.30$ m, so the two torques are $3.0\times0.50 = 1.5$ N $\cdot$ m and $5.0\times0.30 = 1.5$ N $\cdot$ m. The formula also behaves at its edge: letting $a$ grow to $l/2$ drives $F_1$ to zero, and a single force applied at the center of mass is exactly what accelerates a body without turning it.

Remark

The resultant of an antiparallel pair does not lie between the two forces: it lies outside them, beyond the larger one, at a distance $aF_1/(mw)$ past the point where $\vect F_2$ acts. Translation is the demand that this outside point be the center of mass, and that is the whole content of the problem, since it fixes where the middle of the rod is and so how long the rod is. Push the two magnitudes toward equality at fixed $a$ and $F_2$ and the required length grows without bound: in the exact limit the pair is a couple, which has no line of action anywhere and can do nothing but spin the rod.