Kvant Physics Problem 21

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Problem

The armature shafts of two identical direct current electric motors are rigidly connected to each other. If identical current sources with an electromotive force $\mathscr{E}$ are connected to the armature windings, then the angular velocity of the armatures under no-load conditions is equal to $\omega_0$; if the motors are braked so that they do not rotate, then a current $I_0$ flows through the armature windings. One of the sources was reconnected so that the torques of the motors act in opposite directions. What external torque must be applied to the armature shaft so that they rotate with the задан angular velocity $\omega$?

Friction in the motors is negligible; the stator magnetic field is produced by a permanent magnet.

III All-Union Physics Olympiad (1969)

Setup and Assumptions

Two identical direct current motors are rigidly connected by their shafts, so they share the same angular velocity $\omega$ and produce torques that add algebraically.

Each motor is supplied by a source of electromotive force $\mathscr{E}$, with one source reversed so that the motors generate opposing torques.

The armature resistance of each motor is $R$, the torque constant is $k$, and the back electromotive force constant is also $k$, so that the standard relations hold for each motor.

The no-load angular velocity under a single correctly connected source is $\omega_0$, and the current in the blocked rotor case is $I_0$.

The required quantity is the external torque $\tau_{\text{ext}}$ that must be applied to the common shaft so that the system rotates uniformly with angular velocity $\omega$.

Mechanical friction and losses other than Joule heating are neglected, and the magnetic field is assumed constant due to permanent magnets.

Physical Principles

The electrical equation for a DC motor armature is

$\mathscr{E} = IR + k\omega + V_{\text{back}},$

which in standard motor form reduces to

$\mathscr{E} = IR + k\omega.$

The back electromotive force is proportional to angular velocity,

$\mathscr{E}_{\text{back}} = k\omega.$

The electromagnetic torque is proportional to current,

$\tau = kI.$

In steady rotation with constant angular velocity, the total external torque balances the sum of electromagnetic torques,

$\tau_{\text{ext}} + \tau_{\text{em,1}} + \tau_{\text{em,2}} = 0.$

Derivation

For a single motor operating as a generator with angular velocity $\omega$, the current is determined from

$\mathscr{E}_{\text{applied}} - k\omega = IR,$

so

$I = \frac{\mathscr{E}_{\text{applied}} - k\omega}{R}.$

The torque of one motor is therefore

$\tau = k \frac{\mathscr{E}_{\text{applied}} - k\omega}{R}.$

One motor is connected with $\mathscr{E}{1} = +\mathscr{E}$, the other with $\mathscr{E}{2} = -\mathscr{E}$, hence

$\tau_1 = k \frac{\mathscr{E} - k\omega}{R}, \qquad \tau_2 = k \frac{-\mathscr{E} - k\omega}{R}.$

The total electromagnetic torque becomes

$\tau_{\text{em}} = \tau_1 + \tau_2 = k \frac{\mathscr{E} - k\omega - \mathscr{E} - k\omega}{R} = -\frac{2k^2\omega}{R}.$

Uniform rotation requires zero net torque,

$\tau_{\text{ext}} + \tau_{\text{em}} = 0,$

hence

$\tau_{\text{ext}} = \frac{2k^2\omega}{R}.$

The parameters $k$ and $R$ are expressed through the given data. In no-load operation, $I \approx 0$, so

$\mathscr{E} = k\omega_0,$

giving

$k = \frac{\mathscr{E}}{\omega_0}.$

In the blocked state $\omega = 0$, so

$I_0 = \frac{\mathscr{E}}{R},$

which yields

$R = \frac{\mathscr{E}}{I_0}.$

Substituting into the torque expression gives

$\tau_{\text{ext}} = \frac{2}{R}\left(\frac{\mathscr{E}}{\omega_0}\right)^2 \omega = 2 \frac{\mathscr{E}^2 \omega}{\omega_0^2 R}.$

Replacing $R$,

$\tau_{\text{ext}} = 2 \frac{\mathscr{E}^2 \omega}{\omega_0^2} \cdot \frac{I_0}{\mathscr{E}} = 2 \frac{\mathscr{E} I_0 \omega}{\omega_0^2}.$

Result

The required external torque is

$\boxed{\tau_{\text{ext}} = 2 \frac{\mathscr{E} I_0}{\omega_0^2},\omega}.$

In SI units, $\mathscr{E}$ is in volts $\mathrm{V}$, $I_0$ in amperes $\mathrm{A}$, and angular velocity in $\mathrm{rad,s^{-1}}$, so

$\tau_{\text{ext}} = 2 \frac{(\mathrm{V})(\mathrm{A})}{(\mathrm{rad,s^{-1}})^2}(\mathrm{rad,s^{-1}}) = 2,\mathrm{N,m}.$

Thus,

$\boxed{\tau_{\text{ext}} = 2 \frac{\mathscr{E} I_0}{\omega_0^2},\omega \ \mathrm{N,m}}.$

Sanity Checks

Dimensional consistency follows from $\mathscr{E}I_0$ having dimensions of power $\mathrm{W}$, division by $\omega_0^2$ introducing $\mathrm{s^2}$, and multiplication by $\omega$ restoring $\mathrm{N,m}$.

In the limit $\omega \to 0$, the required external torque vanishes, consistent with the absence of back electromotive effects in static conditions.

If $\omega = \omega_0$, the torque becomes $\tau_{\text{ext}} = 2 \mathscr{E} I_0 / \omega_0$, showing linear scaling with speed, as expected for viscous electromagnetic damping.

The sign structure is fixed by the reversed source: both motors contribute torques opposing motion, and any sign error would appear at the step combining $\mathscr{E}$ terms in the summed current expression.