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$A$ paritcle falls freely near the surface of the earth. Consider $a$ fixed point $O$ (not vertically below the particle) on the ground.
Angular momentum of the particle about $O$ is increasing .
The angular velocity of the particle about $O$ is increasing.
The moment of inertia of the particle about $O$ is decreasing .
All of the above
Solution
The magnitude of angular momentum of particle about $\mathrm{O}=\mathrm{mvd}$
since speed $v$ of particle increases, its angular momentum about $0$ increases.
Magnitude of torque of gravitational force about $0=\mathrm{mgd}$
$\Rightarrow$ constant
Moment of inertia of particle about $\mathrm{O}=\mathrm{mr}^{2}$
Hence $Ml$ of particle about $O$ decreases.
angular velocity of particle about $\mathrm{O}=\frac{\mathrm{v} \sin \theta}{\mathrm{r}}$
$\therefore \mathrm{v}$ and $\sin 0$ increase and $\mathrm{r}$ decreases
angular velocity of particle about $O$ increases.
