Kvant Physics Problem 48

A mass $m$ is attached to a rope wound around the shaft of a dynamo armature.

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Problem

F48. A rope is wound around the shaft of the armature of a dynamo machine, and a weight is attached to the rope. As it descends, the weight rotates the armature. When the armature has spun up sufficiently, a load resistance is connected to the machine terminals. Sketch the graph of the armature’s angular velocity as a function of time from the moment the weight begins to move. (X)

G. I. Kosourov

Setup and Assumptions

A mass $m$ is attached to a rope wound around the shaft of a dynamo armature. As the mass descends under gravity, it rotates the armature. The armature and shaft are treated as a rigid body with moment of inertia $I$ about the rotation axis. The shaft has radius $r$, so the angular velocity of the armature is $\omega$ and the linear velocity of the mass is related by

$v = r \omega.$

The goal is to determine the qualitative dependence of $\omega$ on time $t$.

The machine is considered ideal. Mechanical friction and air resistance are neglected. The magnetic field of the dynamo is constant. The electrical load is connected only after the armature has reached a substantial angular velocity. Before the load is connected, the electrical circuit is open and no current flows. After the load is connected, the terminals are connected to a resistance $R$.

The problem requests a sketch of $\omega(t)$ rather than an exact formula.

Physical Principles

The motion of the falling mass and the rotating armature is governed by Newton's second law and the rotational equation of motion. For the falling mass

$mg - T = m a,$

where $T$ is the rope tension and $a$ is the downward acceleration of the mass. For the armature

$T r - M_e = I \alpha,$

where $\alpha$ is the angular acceleration and $M_e$ is any electromagnetic torque acting on the shaft. The kinematic relation between translational and rotational motion is

$a = r \alpha.$

The induced electromotive force of the dynamo is proportional to the angular velocity

$\mathcal E = k \omega,$

where $k$ is a machine constant. When a load resistance $R$ is connected, the current is

$I_e = \frac{\mathcal E}{R} = \frac{k \omega}{R}.$

The electromagnetic torque resisting rotation is proportional to the current

$M_e = k I_e = \frac{k^2}{R} \omega.$

Hence, after the load is connected, the electromagnetic torque is proportional to the angular velocity.

Derivation

Before the load is connected, the circuit is open and $M_e = 0$. The equations of motion are

$mg - T = m a, \qquad T r = I \alpha, \qquad a = r \alpha.$

Substituting $T = I a / r^2$ into the mass equation gives

$mg - \frac{I}{r^2} a = m a.$

Solving for $a$ yields

$a = \frac{mg}{m + I/r^2}.$

This acceleration is constant, and therefore

$\alpha = \frac{a}{r} = \frac{mg}{r (m + I/r^2)}$

is also constant. Starting from rest, the angular velocity grows linearly with time

$\omega(t) = \alpha t.$

The graph of $\omega(t)$ is a straight line with positive slope.

When the load resistance is connected at time $t = t_1$, the electromagnetic torque $M_e = \gamma \omega$ appears, where $\gamma = k^2 / R$. The rotational equation of motion now reads

$T r - \gamma \omega = I \alpha.$

Using the kinematic relation $a = r \alpha$, we can express $\alpha = a / r$ and therefore

$T r - \gamma \omega = I \frac{a}{r} \quad \Rightarrow \quad T = \frac{I}{r^2} a + \frac{\gamma}{r} \omega.$

Substituting this expression for $T$ into the equation for the falling mass yields

$mg - \left(\frac{I}{r^2} a + \frac{\gamma}{r} \omega \right) = m a,$

which simplifies to

$(m + \frac{I}{r^2}) a = mg - \frac{\gamma}{r} \omega.$

Using $a = r \frac{d\omega}{dt}$ gives the linear first-order differential equation

$(m + \frac{I}{r^2}) r \frac{d\omega}{dt} = mg - \frac{\gamma}{r} \omega,$

or equivalently

$\frac{d\omega}{dt} = \frac{mg, r - \gamma \omega}{I + m r^2}.$

This equation governs the angular acceleration after the load is connected. It is linear in $\omega$ and has the standard solution

$\omega(t) = \omega_\infty + \left[\omega(t_1) - \omega_\infty\right] \exp\Bigg(-\frac{\gamma}{I + m r^2}(t - t_1)\Bigg),$

where the terminal angular velocity

$\omega_\infty = \frac{mg, r}{\gamma} = \frac{mg, r, R}{k^2}$

is reached asymptotically as $t \to \infty$. At the instant the load is connected, the angular acceleration changes from

$\alpha_- = \frac{mg}{r (m + I/r^2)}$

to

$\alpha_+ = \frac{mg, r - \gamma \omega(t_1)}{I + m r^2} < \alpha_-,$

so the slope of $\omega(t)$ decreases abruptly. As the system evolves, the angular acceleration diminishes continuously and the angular velocity approaches the constant $\omega_\infty$ smoothly.

Result

Before the load is connected, the angular velocity grows linearly

$\omega(t) = \frac{mg}{r (m + I/r^2)} t.$

After the load is connected, the angular velocity satisfies

$\frac{d\omega}{dt} = \frac{mg, r - \gamma \omega}{I + m r^2},$

approaching the limiting value

$\omega_\infty = \frac{mg, r, R}{k^2}$

exponentially. The required sketch has the qualitative form of a straight line with positive slope transitioning into a smoothly flattening curve that asymptotically reaches a horizontal line.

Sanity Checks

The limiting angular velocity $\omega_\infty$ depends inversely on the resistance $R$, so smaller $R$ produces a larger braking torque and a smaller terminal speed, while larger $R$ reduces braking and increases the terminal speed. When $R \to \infty$, the braking torque vanishes and the angular velocity continues to grow nearly linearly, as expected. The units are consistent throughout: $a$ has units of acceleration, $\alpha$ of angular acceleration, and $M_e = \gamma \omega$ of torque. The instantaneous drop in angular acceleration at $t_1$ is correctly captured by the differential equation, and the smooth approach to $\omega_\infty$ reflects the balance between the decreasing net driving torque and the increasing electromagnetic torque.