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6.System of Particles and Rotational Motion
hard
A particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momnetum of particle about $P$ at $t = \frac{V_0}{g}$ is:-
A
$\frac{1}{2 \sqrt 2} \frac{mV_0^3}{g}$
B
$\frac{1}{2 \sqrt 2} \frac{mV_0^2}{g}$
C
$\frac{1}{2} \frac{mV_0^3}{g}$
D
$\frac{1}{2} \frac{mV_0^2}{g}$
Solution
$\tau=\operatorname{mg} \mathrm{x}=$ torque
$\tau=\operatorname{mg}\left(V_{0} \cos 45\right) t$
$\frac{d J}{d t}=\frac{m g V_{0}}{\sqrt{2}} t$
$\int_{0}^{\mathrm{J}} \mathrm{d} \mathrm{J}=\int_{0}^{\mathrm{t}} \frac{\mathrm{mg} \mathrm{V}_{0}}{\sqrt{2}} \mathrm{t} \mathrm{dt}$
$J=\frac{m g V_{0}}{\sqrt{2}} \frac{t^{2}}{2}$
$J=\frac{m g V_{0}^{3}}{2 \sqrt{2} g^{2}}$
$J=\frac{1}{2 \sqrt{2}} \frac{m V_{0}^{3}}{g}$
Standard 11
Physics
