Tessellate S.T.E.M.S - Physics - School - Set 3 - Problem 5

The rate at which the rod's kinetic energy is lost is equal to the rate at which thermal energy is dissipated in the resistor. Call the induced voltage $\epsilon$ and the bar speed $v$ .

Induced voltage:

$\epsilon = B \, d \, v$

Resistor power:

$P_R = \frac{\epsilon^2}{R} = \frac{B^2 d^2 v^2}{R}$

Bar kinetic energy:

$E = \frac{1}{2} m v^2$

Rate of change of bar kinetic energy:

$\frac{dE}{dt} = \frac{dE}{dv} \frac{dv}{dt} = mv \, \frac{dv}{dt}$

Equating the two:

$P_R = -\frac{dE}{dt} \\ \frac{B^2 d^2 v^2}{R} = -mv \, \frac{dv}{dt} \\ \frac{dv}{dt} = -\frac{B^2 d^2}{m \, R} v$

From here on, the problem can be easily solved numerically. It can also be solved analytically, as desired.

Addendum (analytical solution):

We can remove time from the equation by applying the chain rule to our result.

$\frac{dv}{dt} = -\frac{B^2 d^2}{m \, R} v \\ \frac{dv}{dx} \frac{dx}{dt} = -\frac{B^2 d^2}{m \, R} \frac{dx}{dt} \\ \frac{dv}{dx} = -\frac{B^2 d^2}{m \, R} \\ dx = -\frac{m \, R}{B^2 d^2} \, dv$

Integrate to find the total displacement:

$\Delta x = -\frac{m \, R}{B^2 d^2} \int_{v_0}^0 \, dv = \frac{m \, R}{B^2 d^2} \int_0^{v_0} \, dv \\ = \frac{m \, R \, v_0}{B^2 d^2}$