Question

A stone is dropped from point A, which is at a horizontal distance $l$ from point 'O'. When the stone has fallen a depth '$h$'

• A. Angular momentum of the stone about the point 'O' is $ml\sqrt{2gh}$
• B. Angular velocity of the stone about the point 'O' is $\frac{\sqrt{2gh}}{l}$
• C. Angular velocity of the stone about the point 'O' is $\frac{l}{l^2+h^2}\sqrt{2gh}$
• D. Net torque on the stone about the point 'O' is $\left(mg\right)\left(\sqrt{l^2+h^2}\right)$

Answer 1

Answer: (A), (C)

The correct options are
A Angular momentum of the stone about the point 'O' is $ml\sqrt{2gh}$
C Angular velocity of the stone about the point 'O' is $\frac{l}{l^2+h^2}\sqrt{2gh}$

$L=mv{r}_1=m\left(\sqrt{2gh}\right)\left(l\right)$ (Where, $r_1$ is perpendicular distance from velocity vector which is $l$ and $v=\sqrt{2gh}$)


(A) is correct

$tan\theta =\frac{y}{l}$

Differerntiating both sides,

$se{c}^2\theta \frac{d\theta }{dt}=\frac{1}{l}\left(\frac{dy}{dt}\right)$

$\omega =\frac{d\theta }{dt}=\frac{1}{lse{c}^2\theta }\left(\sqrt{2gh}\right)$

$(\because \frac{dy}{dt}=v=\sqrt{2gh})$

$\therefore \omega =\frac{\sqrt{2gh}}{l(1+ta{n}^2\theta )}=\frac{\sqrt{2gh}}{l(1+{\left(\frac{h}{l}\right)}^2)}=\frac{l}{l^2+h^2}\sqrt{2gh}$

(B) is wrong. (C) is correct.

Torque about O,
$\tau =F{r}_1=\left(mg\right)\left(l\right)$ (Where, $r_1$ is perpendicular distance from force vector which is $l$)
(D) is wrong.