Question

A rod of length $L=\frac{10}{3}m$ and density $\rho$ is pivoted about its one end is submerged in liquid of density $2\rho$ and held horizontal as shown. Now if the rod is released from rest, the angular speed in $rad/s$ when it becomes vertical is [Answer upto two decimal points][Take $g=10m/s^2$]

Answer 1

As the density of rod is less than the liquid, the buoyancy force $B$ will be acting upwards on the rod and the weight of the rod acts downwards. Let $M$, $V$ be the mass and volume of the rod respectively.

Hence using the work-energy theorem.
$\Rightarrow B\frac{L}{2}-Mg\frac{L}{2}=\frac{1}{2}{I}_o{\omega }^2$
where $I_o$ is the moment of Inertia
$\text{about}O,I_o=\frac{1}{3}M{L}^2\Rightarrow \left(2\rho Vg\right)\frac{L}{2}-\left(\rho Vg\right)\frac{L}{2}=\frac{1}{2}\frac{M}{3}{L}^2{\omega }^2\Rightarrow V\rho g\frac{L}{2}=\frac{1}{2}\frac{\left(\rho V\right)}{3}{L}^2{\omega }^2\Rightarrow \omega =\sqrt{\frac{3g}{L}}=\sqrt{\frac{3\times 10\times 3}{10}}=3rad/s$