- #1
Toby_phys
- 26
- 0
Homework Statement
A homopolar generator consists of a metal disc of radius ##a## and a central axle which has radius ##a/4##. The disc has resistivity ##\rho## and thickness ##t##. It is rotated in a uniform magnetic field ##B## about an axis through the centre, which is parallel to ##B## and perpendicular to the plane containing the disc, at an angular frequency ##\omega##. Thin ring brushes make good electrical contact with the disc near the axle and near the outer rim of the disc as shown.
(a) Calculate the resistance of the disc ##R_D## measured between the brushes.
(b) Show that the potential difference between the brushes is ##(15/32)ωBa2## .
(c) A load resistance ##R_L## is connected across the generator and the drive is removed. Show that, in the absence of mechanical friction, the time ##\tau## taken for the disc to slow down to half its initial angular speed is
$$\tau =( 32/15)^2 \frac{m(R_L + R_D) ln(2)}{ 2a^2B^2} $$.
Homework Equations
Inertia of a disc:
$$I=1/2 ma^2$$
Kinetic energy of rotating disc:
$$E=1/2 I\dot{\theta}^2$$
Electrical Power Dissipated:
$$ P=\frac{dE}{dt}=\frac{v^2}{R_l+R_D}=(\frac{15Ba^2}{32})^2\frac{\dot{\theta}^2}{R_l+R_D}$$
The Attempt at a Solution
I have done part (a) and (b) - its only part C I need:
Energy at full speed:
$$E_1=\frac{ma^2\omega^2}{4}$$
Energy at half speed:
$$E_2=\frac{ma^2\omega^2}{16}$$
Energy lost:
$$E_2-E_1=\frac{3}{16}ma^2\omega^2=\int^\tau_0pdt=(\frac{15Ba^2}{32})^2\int^\tau_0\frac{\dot{\theta}^2}{R_L+R_D}dt$$
Basically, how do I solve that integral? I know the start and end values of ##\dot{\theta}##/
thank you