Question

A conical pendulum consists of a bob of mass $m$ in motion in a circular path in a horizontal plane as shown in figure. During the motion, the supporting wire of length $l$, maintains a constant angle $\theta$ with the vertical. The magnitude of the angular momentum of the bob about the vertical dashed line is

• A. ${\left(\frac{m^{2}g{l}^3{\cos }^4\theta }{\sin \theta }\right)}^{1/2}$
• B. ${\left(\frac{m^{2}g{l}^3{\sin }^4\theta }{\cos \theta }\right)}^{1/2}$
• C. ${\left(\frac{m^{2}g{l}^3{\sin }^3\theta }{\cos \theta }\right)}^{1/2}$
• D. ${\left(\frac{m^{2}g{l}^3{\cos }^3\theta }{\sin \theta }\right)}^{1/2}$

Answer 1

Answer: (B)

We start with the particle under a net force model in the $x$ and $y$ directions:

$\sum F_x=m{a}_x:T\sin \theta =\frac{m{v}^2}{r}$
$\sum F_y=m{a}_y:T\cos \theta =mg$
So $\frac{\sin \theta }{\cos \theta }=\frac{v^2}{rg}$ and $v=\sqrt{rg\frac{\sin \theta }{\cos \theta }}$
then $L=rmv\sin 90^{\circ }=rm\sqrt{rg\frac{\sin \theta }{\cos \theta }}=\sqrt{m^{2}g{r}^3\frac{\sin \theta }{\cos \theta }}$
and since $r=l\sin \theta$,
$L=\sqrt{m^{2}g{l}^3\frac{{\sin }^4\theta }{\cos \theta }}$