Irodov 3.295: Steering a Rod Against Gravity – Amit & Deepali Sinha Foundation

Problem Irodov 3.295

A metal rod of mass $m$ is pivoted at one end on a horizontal axis $O$ and turns in a vertical plane, its other end sliding along a circular conducting ring of radius $a$ centered on $O$ . A uniform magnetic field of induction $B$ is perpendicular to the plane of the ring (into the page as drawn). A source of emf is connected between the axis and the ring, forming a circuit of total resistance $R$ . The rod turns clockwise, and $\omega t$ is its angle from the upward vertical. Neglecting friction, the inductance of the circuit and the resistance of the ring, find how the emf of the source must vary in time for the rod to turn with a constant angular velocity $\omega$ .

The ring of radius a and the rod pivoted at its center O on a horizontal axis, turning clockwise, at the angle ω t from the upward vertical; the source joined to O and to the ring, and the field B into the page. — The ring of radius $a$ and the rod pivoted at its center $O$ on a horizontal axis, turning clockwise, at the angle $\omega t$ from the upward vertical; the source joined to $O$ and to the ring, and the field $\vect B$ into the page.

Given

$m$
$a$
$B$ normal to the ring
$R$ (whole circuit)
$\omega$ constant
$\omega t$ from the upward vertical, clockwise
uniform rod (assumed)
no friction

Find

\mathcal{E}(t)

of the source

Solution

Step 1The rod’s own emf

The rod turns about $O$ , so an element $dr$ at distance $r$ moves at $\omega r$ , perpendicular to the rod and to $\vect B$ . Summing the motional emfs of the elements from the axis to the ring:

\mathcal{E}_m = \int_0^a \omega rB\,dr = \frac{\omega Ba^2}{2} \tag{1}

It is constant, since $\omega$ is. On its own it would drive a current along the rod whose Ampere force opposes the rotation (Lenz).

Step 2The torque of the Ampere force

Let $I$ be the current along the rod, counted positive in the sense $\mathcal{E}_m$ drives it. The force on an element $dr$ is $IB\,dr$ , perpendicular to the rod, and for $I > 0$ it opposes the rotation. Its torque about $O$ is

N_A = \int_0^a IBr\,dr = \frac{IBa^2}{2} \tag{2}

During the first half turn: the weight m g at the rod's middle turns it on, clockwise; the Ampere force F of the current I along the rod holds it back. — During the first half turn: the weight $m\vect g$ at the rod’s middle turns it on, clockwise; the Ampere force $\vect F$ of the current $I$ along the rod holds it back.

Step 3The torque balance

The weight acts at the rod’s middle (a uniform rod, assumed), a distance $a/2$ from $O$ ; its lever arm is $(a/2)\sin\omega t$ , and while $0 < \omega t < \pi$ it turns the rod on, in the sense of rotation. With no friction and $\omega$ constant, the net torque about $O$ must vanish; using (2):

\frac{IBa^2}{2} = mg\,\frac{a}{2}\sin\omega t

I = \frac{mg}{aB}\sin\omega t

(3)

For $\pi < \omega t < 2\pi$ the rod rises, the weight holds it back, and $I$ is negative: the current reverses and the Ampere force pushes the rod on.

Step 4The emf the source must supply

Count the source’s emf $\mathcal{E}$ positive in the sense that drives the current against $\mathcal{E}_m$ , the sense in which the source turns the rod forward. Kirchhoff’s loop law round the circuit of resistance $R$ then reads

\mathcal{E}_m - \mathcal{E} = IR

Solving for $\mathcal{E}$ and substituting (1) and (3):

\mathcal{E}(t) = \frac{\omega Ba^2}{2} - \frac{mgR}{aB}\sin\omega t

(4)

The constant part cancels the rod’s own emf, so that without gravity no current would flow and no torque act; the varying part drives the current that balances the weight’s torque. The sign of the sine term is set by where $\omega t$ is measured from: from the upward vertical, as here, the rod first falls and gravity helps, so the source must hold it back; with $\omega t$ measured from the downward vertical the rod would first rise and the term would change sign. A mirror image of the whole arrangement (counter-clockwise rotation, $\omega t$ measured counter-clockwise from the upward vertical) leaves the formula unchanged.

Check

Limits. A weightless rod, $m \to 0$ , gives $\mathcal{E} = \omega Ba^2/2$ : the source exactly cancels the rod’s own emf, no current flows, no torque acts, and a free rod indeed keeps turning at constant $\omega$ . At $\omega t = \pi/2$ , the rod horizontal on the side it falls toward, the weight’s torque is largest and helping, and (4) is smallest, $\omega Ba^2/2 - mgR/(aB)$ : the source backs off most where the most braking current is needed. At $\omega t = 0$ and $\pi$ , the rod vertical, the weight has no torque and $\mathcal{E}$ is again the rod’s own emf. Dimensions: $mgR/(aB)$ is $\text{N}\cdot\Omega/(\text{m}\cdot\text{T}) = \text{A}\cdot\Omega = \text{V}$ , since $\text{T}\cdot\text{m} = \text{N/A}$ .

Remark

Averaged over a turn the sine term vanishes, and the source’s mean emf is just the rod’s own emf, $\omega Ba^2/2$ . Its mean power is not zero, though: gravity does no net work over a turn, but the current that balances it heats the circuit, and with the mean of $\sin^2\omega t$ equal to $1/2$ the source pays on average

\langle I^2R\rangle = \frac{m^2g^2R}{2a^2B^2}

A stronger field makes the control cheaper: the correction to the emf falls as $1/B$ and this loss as $1/B^2$ .