43. Terminal Velocity of Rolling Cylinder
Step 1: Energy Balance Approach
At terminal velocity $v$, the cylinder rolls without accelerating. The rate of potential energy loss (Gravity Power) equals the rate of electrical energy dissipation (Resistive Heating).
$$ P_{gravity} = P_{electrical} $$Step 2: Gravitational Power
The component of force down the incline is $mg \sin \theta$.
$$ P_{gravity} = F \cdot v = (mg \sin \theta) v $$Step 3: Electrical Power
The coil rotates in a vertical magnetic field $B$. The flux through the coil is $\Phi = B A \cos(\omega t)$, where $A = 2r \cdot l$ and $\omega = v/r$.
Induced EMF: $\varepsilon = -\frac{d\Phi}{dt} = B A \omega \sin(\omega t)$.
Average electrical power dissipated:
$$ \langle P_{elec} \rangle = \frac{\langle \varepsilon^2 \rangle}{R} = \frac{B^2 A^2 \omega^2}{R} \langle \sin^2(\omega t) \rangle = \frac{B^2 A^2 \omega^2}{2R} $$Step 4: Solve for Velocity
Equating the powers:
$$ (mg \sin \theta) v = \frac{B^2 (2rl)^2 (v/r)^2}{2R} $$ $$ mg v \sin \theta = \frac{B^2 (4r^2 l^2) (v^2/r^2)}{2R} $$ $$ mg v \sin \theta = \frac{2 B^2 l^2 v^2}{R} $$Cancel $v$ and solve:
$$ v = \frac{R mg \sin \theta}{2 B^2 l^2} $$
$$ v = \frac{R mg \sin \theta}{2 B^2 l^2} $$