Question

In above figure, a laser beam of power $4.60W$ and diameter $D=2.60mm$ is directed upward at one circular face (of diameter $d<2.60mm$) of a perfectly reflecting cylinder. The cylinder is levitated because the upward radiation force matches the downward gravitational force. If the cylinder’s density is $1.20g/c{m}^3$, what is its height $H$?

Answer 1

The mass of the cylinder is $m=\rho \left(\pi D^2/4\right)H$, where $D$ is the diameter of the cylinder.

Since it is in equilibrium $F_{net}=0$

$F_{net}=mg-F_r=\frac{\pi H{D}^{2}g\rho }{4}-\left(\frac{\pi D^2}{4}\right)\left(\frac{2I}{c}\right)=0$.
We solve for $H$:
$H=\frac{2I}{gc\rho }=\left(\frac{2P}{\pi D^2/4}\right)\frac{1}{gc\rho }$,
$=\frac{2\left(4.60W\right)}{\left[\pi {(2.60\ast {10}^{-3}m)}^2/4\right]\left(9.8m/s^2\right)(3.0\ast {10}^{8}m/s)(1.20\ast {10}^{3}kg/m^3)}=4.91\times {10}^{-7}m$.