A different wheel

We proceed as follows..... Let the current in the circuit at an instant be $i$ .Then using Kirchoff's Loop Law. $Ldi/dt+E-V=0$ . $E$ is the magnitude of the motional Emf induced in the spoke.....We know $E=B\omega R^2/2$ .Also $Id\omega /dt=\tau$ .For a small element $d\tau=(iBdr)r$ .Now integration with limits from $0-R$ yield $\int d\tau=iB\int rdr$ or $\tau =iBR^2/2$ .Plugging the required eq becomes $iBR^2/2=Id\omega /dt$ .So Our KVL modifies to $2LI/BR^2d\omega /dt=E-B\omega R^2/2$ .Now its easy to observe that the eq represents the equation of SHM(set $E-B\omega R^2/2=\phi$ you get the trivial eq $d^2\phi /dt^2=-\psi^2 \phi$ ).Now we use the boundary conditions. 1)At $t=0,i=0$ .& 2) $\omega =0$ .This gives $\omega=2V/BR^2(1-cos\psi t)$ .where $\psi=BR^2/2√(IL)$ .Plugging the required values you get the answer as $\omega =5$

The trick to this problem, which I initially didn't fully comprehend, is that the charges in the spoke (rod) effectively have two different motions:

1) A longitudinal electrical drift which results in a transverse torque on the spoke (via the B-field)
2) A transverse motion, due to the rotation, which induces a longitudinal motional emf in the rod (also via the B-field).

When the current and B-field are at right angles, the transverse force on a current carrying wire of length $l$ is:

$\large{F = ilB}$

The differential torque on a piece of the conducting spoke is: $\large{d\tau = (i \, dr\, B)r = B i r \, dr}$

The total torque on the spoke is:

$\large{\tau = \int_0^R B i r \, dr = \frac{B i R^2}{2}}$

Applying the rotational form of Newton's Second Law gives:

$\large{\tau = I \frac{d\omega}{dt} = \frac{B i R^2}{2} \\ i = \frac{2I}{B R^2} \frac{d\omega}{dt} \\ \frac{di}{dt} = \frac{2I}{B R^2} \frac{d^2 \omega}{dt^2}}$

There are three voltage changes in the circuit:

1) The source voltage
2) The inductor voltage $V_L = L \frac{di}{dt}$
3) The motional emf induced in the rod

Let's calculate the motional emf. For a rod of length $l$ moving at speed $v$ at right angles to a uniform B-field, the motional emf is:

$\large{E = Blv}$

For an infinitesimal piece of the spoke, this translates to:

$\large{dE = B \, dr \, r \omega = B \omega r \, dr}$

The total motional emf in the spoke is then:

$\large{E = \int_0^R B \omega r \, dr = \frac{B \omega R^2}{2}}$

Writing the KVL equation for the circuit gives:

$\large{V = L \frac{di}{dt} + \frac{B \omega R^2}{2}}$

Substituting for $\frac{di}{dt}$ in terms of $\omega$ gives:

$\large{V = \frac{2IL}{B R^2} \frac{d^2 \omega}{dt^2} + \frac{B \omega R^2}{2}}$

Do these sign conventions make sense? It seems to me that this system will likely behave similarly to an un-commutated DC motor, which means that its angular speed will oscillate between clockwise and counter-clockwise when a DC voltage is applied. This behavior is implied by the above equations.

The homogeneous equation is:

$\large{\frac{d^2 \omega}{dt^2} = -\frac{B^2 R^4}{4IL} \omega = -\alpha \omega}$

The general solution is then:

$\large{\omega = A cos(\sqrt{\alpha} t) + B sin(\sqrt{\alpha} t) + C \\ \dot{\omega} = -A \sqrt{\alpha} sin(\sqrt{\alpha} t) + B \sqrt{\alpha} cos(\sqrt{\alpha} t) \\ \ddot{\omega} = -A \alpha cos(\sqrt{\alpha} t) - B \alpha sin(\sqrt{\alpha} t)}$

Applying initial conditions gives (note that the initial speed is zero and the initial torque is zero):

$\large{0 = A + C \\ 0 = B \\ V = \frac{2IL}{B R^2} (-A \alpha) = \frac{2IL}{B R^2} (-A) \frac{B^2 R^4}{4IL} = -A \frac{B R^2}{2} }$

Solving for the constants gives:

$\large{A = -\frac{2V}{BR^2} \\ B = 0 \\ C = \frac{2V}{BR^2}}$

The expression for $\omega$ is then:

$\large{\omega = \frac{2V}{BR^2} (1 - cos(\sqrt{\alpha} t)) \\ \alpha = \frac{B^2 R^4}{4 IL} }$

Plugging in numbers and solving at $\large{t = \frac{2 \pi}{3}}$ gives $\boxed{\large{\omega = 5}}$ .