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(C) The only force acting on the projectile is its weight $mg$ acting downwards.Taking point $P$ as the origin,the torque $	au$ due to gravity at any

Vedclass · Accepted Answer

$\frac{1}{2\sqrt{2}} \cdot \frac{mv_0^3}{g}$

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(C) The only force acting on the projectile is its weight $mg$ acting downwards.Taking point $P$ as the origin,the torque $	au$ due to gravity at any time $t$ is given by $	au = \vec{r} 	imes \vec{F} = x(t) \cdot mg$,where $x(t) = (v_0 \cos 45^{\circ})t = \frac{v_0}{\sqrt{2}}t$.Thus,$	au = mg \cdot \frac{v_0}{\sqrt{2}}t$.The change in angular momentum $\Delta L$ is equal to the angular impulse:$\Delta L = \int_{0}^{t} 	au \, dt = \int_{0}^{t} \frac{mgv_0}{\sqrt{2}} t \, dt = \frac{mgv_0}{\sqrt{2}} \left[ \frac{t^2}{2} ight]_{0}^{t} = \frac{mgv_0 t^2}{2\sqrt{2}}$.Since the initial angular momentum at $P$ is zero (the line of action of the initial velocity passes through $P$),the angular momentum at time $t = \frac{v_0}{g}$ is:$L = \frac{mgv_0}{2\sqrt{2}} \left( \frac{v_0}{g} ight)^2 = \frac{mgv_0^3}{2\sqrt{2}g^2} = \frac{mv_0^3}{2\sqrt{2}g}$.

$A$ particle of mass $m$ is projected from a point $P$ on the ground with an initial velocity $v_0$ at an angle of $45^{\circ}$ with the horizontal at $t = 0$. Find the magnitude of the angular momentum of the particle at time $t = \frac{v_0}{g}$.

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