Kvant Physics Problem 21
Two identical DC motors have their armature shafts rigidly connected, so both armatures rotate with the same angular velocity $\omega$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m45s
Source on kvant.digital
Problem
The armature shafts of two identical DC motors are rigidly connected to each other. If identical power sources with emf $\mathscr{E}$ are connected to the armature windings, then the no-load angular velocity of the armatures is $\omega_0$; if the motors are braked so that they do not rotate, a current $I_0$ flows through the armature windings. One of the power sources is then switched so that the torques produced by the motors are opposite in direction. What external torque must be applied to the armature shaft so that it rotates with a prescribed angular velocity $\omega$? Friction in the motors is negligibly small; the stator magnetic field is produced by a permanent magnet.
III All-Union Physics Olympiad (1969)
Setup and Assumptions
Two identical DC motors have their armature shafts rigidly connected, so both armatures rotate with the same angular velocity $\omega$. The motors are supplied by identical sources of electromotive force $\mathscr E$.
When a single motor runs without mechanical load, its angular velocity is $\omega_0$. When its shaft is blocked, the armature current is $I_0$.
One source is then reversed, so that the electromagnetic torques produced by the two motors act in opposite directions. The shaft is required to rotate steadily with prescribed angular velocity $\omega$. The unknown quantity is the external torque $M_{\text{ext}}$ that must be applied to the common shaft.
The stator field is produced by permanent magnets, so the magnetic flux is constant. Friction and other mechanical losses are neglected. The motors are identical, and the standard linear relations for a DC motor are assumed to hold.
Physical Principles
For a DC motor with constant magnetic flux, the back electromotive force is proportional to angular velocity:
$$E_b = k\omega,$$
where $k$ is a motor constant.
The electromagnetic torque is proportional to armature current:
$$M = kI.$$
The same constant $k$ appears in both relations when SI units are used.
The armature circuit obeys Ohm's law:
$$I=\frac{U-E_b}{R},$$
where $U$ is the applied voltage and $R$ is the armature resistance.
For steady rotation with constant angular velocity, the sum of torques acting on the shaft must vanish:
$$M_{\text{ext}}+M_1+M_2=0.$$
Derivation
First determine the motor constants from the data given.
When the motor is blocked,
$$\omega=0,$$
hence the back emf vanishes:
$$E_b=0.$$
The current is then
$$I_0=\frac{\mathscr E}{R},$$
so
$$R=\frac{\mathscr E}{I_0}.$$
Next consider no-load operation. Since friction is neglected, the current tends to zero in the no-load state. Consequently the voltage drop across the armature resistance is zero, and
$$E_b=\mathscr E.$$
Using $E_b=k\omega$,
$$k\omega_0=\mathscr E,$$
hence
$$k=\frac{\mathscr E}{\omega_0}.$$
Now analyze the situation with one source reversed.
Let motor 1 be connected normally. Its applied voltage is
$$U_1=\mathscr E.$$
Its current is
$$I_1=\frac{\mathscr E-k\omega}{R}.$$
Its electromagnetic torque is
$$M_1=kI_1 =\frac{k}{R}\left(\mathscr E-k\omega\right).$$
Motor 2 has its source reversed, so
$$U_2=-\mathscr E.$$
Its current is
$$I_2=\frac{-\mathscr E-k\omega}{R}.$$
The corresponding torque is
$$M_2=kI_2 =\frac{k}{R}\left(-\mathscr E-k\omega\right).$$
The total electromagnetic torque acting on the shaft is
$$M_1+M_2 = \frac{k}{R} \Big[ (\mathscr E-k\omega) + (-\mathscr E-k\omega) \Big].$$
The terms containing $\mathscr E$ cancel:
$$M_1+M_2 = -\frac{2k^2\omega}{R}.$$
For steady rotation,
$$M_{\text{ext}}+M_1+M_2=0,$$
therefore
$$M_{\text{ext}} = \frac{2k^2\omega}{R}.$$
Substituting
$$k=\frac{\mathscr E}{\omega_0}, \qquad R=\frac{\mathscr E}{I_0},$$
gives
$$M_{\text{ext}} = 2 \frac{\left(\frac{\mathscr E}{\omega_0}\right)^2} {\frac{\mathscr E}{I_0}} ,\omega.$$
After simplification,
$$M_{\text{ext}} = \frac{2\mathscr E I_0}{\omega_0^2},\omega.$$
Result
The external torque required to maintain the prescribed angular velocity $\omega$ is
$$M_{\text{ext}} = \frac{2\mathscr E I_0}{\omega_0^2},\omega.$$
Equivalently,
$$\boxed{ M_{\text{ext}} = 2\left(\frac{\mathscr E}{\omega_0}\right) \left(\frac{I_0}{\omega_0}\right)\omega }.$$
No numerical values are provided in the statement, so the answer remains in symbolic form:
$$\boxed{ M_{\text{ext}} = \frac{2\mathscr E I_0}{\omega_0^2},\omega } \qquad (\text{N}\cdot\text{m}).$$
Sanity Checks
The combination $\mathscr E I_0$ has units of power:
$$[\mathscr E I_0] = \mathrm V\cdot\mathrm A = \mathrm W.$$
Hence
$$\left[ \frac{\mathscr E I_0}{\omega_0^2}\omega \right] = \frac{\mathrm W}{(\mathrm s^{-1})^2},\mathrm s^{-1} = \mathrm W,\mathrm s = \mathrm N,\mathrm m,$$
which is the correct unit for torque.
For $\omega=0$,
$$M_{\text{ext}}=0.$$
At rest the torques produced by the two motors are equal and opposite, so no external torque is needed.
The required torque is proportional to $\omega$. This is expected because the reversed connection makes the pair behave like an electromagnetic brake whose braking torque grows with rotational speed.
The most common sign error occurs when writing the current in the reversed motor. Its applied voltage is $-\mathscr E$, giving
$$I_2=\frac{-\mathscr E-k\omega}{R},$$
not
$$\frac{-\mathscr E+k\omega}{R}.$$
Using the wrong sign would prevent the cancellation of the $\mathscr E$ terms and lead to an incorrect torque expression.