Irodov 3.297: A Capacitor Adds to the Mass – Amit & Deepali Sinha Foundation

Problem Irodov 3.297

Two smooth parallel metal bars a distance $l$ apart lie in a plane inclined at an angle $\alpha$ to the horizontal, running down the slope, and are joined at their upper ends through a capacitor of capacitance $C$ . A metal rod of mass $m$ lies across the bars and slides down them under gravity, keeping contact with both. A uniform magnetic field of induction $B$ is directed perpendicular to the inclined plane. The resistances of the bars, the rod and the contacts are negligible, and so is the self-inductance of the circuit. Find the acceleration of the rod.

The bars inclined at α, joined at the top through the capacitor C, with the connector of mass m sliding down at v under its weight m g. The field B is normal to the bars' plane. — The bars inclined at $\alpha$ , joined at the top through the capacitor $C$ , with the connector of mass $m$ sliding down at $\vect v$ under its weight $m\vect g$ . The field $\vect B$ is normal to the bars’ plane.

Given

$m$
$\alpha$
$C$
$l$
$B$ normal to the plane
smooth bars
no resistance
no self-inductance

Find

the acceleration

w

Solution

Step 1The capacitor follows the emf

The connector moving at $\vect v$ across the field has the motional emf $Blv$ , as in the resistor version. With no resistance and no inductance anywhere in the circuit, the capacitor’s voltage equals that emf at every instant, so its charge is

q = C\,Blv \tag{1}

Step 2The current is set by the acceleration

As the connector speeds up the charge (1) grows, and the current in the circuit is its rate of change. Writing $w = dv/dt$ for the acceleration:

I = \frac{dq}{dt} = CBl\,\frac{dv}{dt} = CBl\,w \tag{2}

While the connector accelerates, the charging current I flows round the circuit, the plate joined to the connector's + end charging positive, and the Ampere force F on the connector points up the slope. — While the connector accelerates, the charging current $I$ flows round the circuit, the plate joined to the connector’s $+$ end charging positive, and the Ampere force $\vect F$ on the connector points up the slope.

Step 3The magnetic brake

The current (2) in the connector sits in the field, which is perpendicular to it, so the connector feels the Ampere force along the bars; by Lenz’s rule it opposes the motion, up the slope:

F = IlB = CB^2l^2\,w \tag{3}

Step 4Newton’s law along the slope

Along the bars act the weight’s component $mg\sin\alpha$ down the slope and the force (3) up it; the normal reaction of the smooth bars and the weight’s normal component balance. Taking down the slope as positive:

mw = mg\sin\alpha - CB^2l^2\,w

Collecting the terms in $w$ and solving:

w = \frac{g\sin\alpha}{1 + CB^2l^2/m}

(4)

The acceleration is constant: the connector slides with uniform acceleration, less than $g\sin\alpha$ , and never reaches a terminal speed.

Check

Energy gives the same result. The weight’s power $mg\sin\alpha\,v$ goes into the connector’s kinetic energy and the capacitor’s energy, since nothing dissipates:

mg\sin\alpha\,v = \frac{d}{dt}\left(\frac{mv^2}{2} + \frac{C(Blv)^2}{2}\right) = \left(m + CB^2l^2\right)v\,w,

whose solution for $w$ is (4). Limits: $C \to 0$ (an open circuit) gives the free slide, $w = g\sin\alpha$ ; dimensions: $CB^2l^2$ is $(\text{C/V})\cdot(\text{V}\cdot\text{s/m}^2)^2\cdot\text{m}^2 = \text{C}\cdot\text{V}\cdot\text{s}^2/\text{m}^2 = \text{J}\cdot\text{s}^2/\text{m}^2 = \text{kg}$ , a mass.

Remark

The capacitor makes the connector behave as if its mass were $m + CB^2l^2$ , while gravity still pulls only on $m$ : the field couples the rod’s motion to the capacitor’s charge, and the capacitor’s energy $C(Blv)^2/2$ plays the part of extra kinetic energy. The charging current (2) is constant, so the charge grows in step with the speed, linearly in time.