Question

Two small balls of mass $m=2\text{kg}$ each are attached to a light rod of length $l=2\text{m}$, one at its centre and the other at a free end. The rod is fixed at the other end and is rotated in the horizontal plane at an angular speed $\omega =5\text{rad/s}$. Calculate the angular momentum of the ball at the end with respect to the ball at the centre in ${\text{kg m}}^2\text{/s}$

Answer 1

As we know, $V=\omega R$
The velocity of the ball $A$ with respect to the fixed end is $V_A=\omega \left(\frac{l}{2}\right)=\frac{\omega l}{2}$ and that of $B$ with respect to fixed end is $V_B=\omega l$
Hence, the angular momentum of $B$ with respect to $A$ is
$L_{BA}=m{V}_{BA}{r}_{BA}=m(V_B-V_A)\left(\frac{l}{2}\right)$
$=m\times (\omega l-\frac{\omega l}{2})\times \left(\frac{l}{2}\right)$
$=2\times \left(\frac{5\times 2}{2}\right)\times \left(\frac{2}{2}\right)$
$\therefore L_{BA}=10{\text{kg m}}^2\text{/s}$