Question

A uniform rod of mass $m$ and length $l=1m$ is hinged to a cart in vertical position as shown in figure. If cart starts moving at acceleration $a=2m/s^2$. EMF induced(in volts) across rod when it becomes horizontal, if uniform magnetic field of B = 2T exist in space is $[g=10m/s^2]$

Answer 1

When the frame of reference is connected to the truck, there will be a pseudo force of $ma$ acting on the rod at the centre, in the backward direction.
Work done by the pesudo torque and the weight as the rod rotates by $d\theta$ is,
$dW={\int }_0^{\frac{\pi }{2}}ma\frac{l}{2}\cos \theta d\theta +mg\frac{l}{2}\sin \theta d\theta =mg\frac{L}{2}+ma\frac{L}{2}$

By work energy theorem,
$mg\frac{L}{2}+ma\frac{L}{2}=\frac{1}{2}I{\omega }^2$
${\omega }^2=\frac{3(g+a)}{L}$
$\epsilon =\frac{B\omega L^2}{2}=\frac{B}{2}{L}^2\sqrt{\frac{3(g+a)}{L}}=\sqrt{36}$
$\epsilon =6.708$ V