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:

$\sum F_{x}=m a_{x}: T \sin \theta=\frac{m v^{2}}{r}$
$\sum F_{y}=m a_{y}: T \cos \theta=m g$
So $\frac{\sin \theta}{\cos \theta}=\frac{v^{2}}{r g}$ and $v=\sqrt{r g \frac{\sin \theta}{\cos \theta}}$
then $L=r m v \sin 90^{\circ}=r m \sqrt{r g \frac{\sin \theta}{\cos \theta}}=\sqrt{m^{2} g r^{3} \frac{\sin \theta}{\cos \theta}}$
and since $r=l \sin \theta$,
$L=\sqrt{m^{2} g l^{3} \frac{\sin ^{4} \theta}{\cos \theta}}$