Answer
$v = 1.46~m/s$
Work Step by Step
We can find the moment of inertia of the rod and spheres.
$I = \frac{1}{12}(0.120~kg)(0.800~m)^2+(0.0200~kg)(0.400~m)^2+(0.0500~kg)(0.400~m)^2$
$I = 0.0176~kg~m^2$
We can use conservation of energy to find the angular speed. The kinetic energy will be equal in magnitude to the change in potential energy.
$KE = PE$
$\frac{1}{2}I\omega^2 = (0.0500~kg)(9.80~m/s^2)(0.400~m)- (0.0200~kg)(9.80~m/s^2)(0.400~m)$
$\omega^2 = \frac{(2)(0.0300~kg)(9.80~m/s^2)(0.400~m)}{I}$
$\omega = \sqrt{\frac{(2)(0.0300~kg)(9.80~m/s^2)(0.400~m)}{0.0176~kg~m^2}}$
$\omega = 3.656~rad/s$
We can find the linear speed.
$v = \omega ~R = (3.656~rad/s)(0.400~m)$
$v = 1.46~m/s$
