A particle of mass m is projected at 45^o at | vedclass

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Question

(A) The torque $	au$ acting on the particle about point $P$ is due to the gravitational force $mg$ acting downwards at a horizontal distance $x$ from

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}} \frac{mV_0^3}{g}$

Vedclass · Answer

(A) The torque $	au$ acting on the particle about point $P$ is due to the gravitational force $mg$ acting downwards at a horizontal distance $x$ from $P$.$	au = mgx$Since the horizontal velocity is constant,$x = (V_0 \cos 45^o) t = \frac{V_0}{\sqrt{2}} t$.Therefore,$	au = mg \left( \frac{V_0}{\sqrt{2}} t ight) = \frac{mgV_0}{\sqrt{2}} t$.We know that torque is the rate of change of angular momentum: $	au = \frac{dL}{dt}$.Integrating with respect to time from $t = 0$ to $t = \frac{V_0}{g}$:$L = \int_{0}^{t} 	au dt = \int_{0}^{V_0/g} \frac{mgV_0}{\sqrt{2}} t dt$$L = \frac{mgV_0}{\sqrt{2}} \left[ \frac{t^2}{2} ight]_{0}^{V_0/g}$$L = \frac{mgV_0}{\sqrt{2}} \cdot \frac{1}{2} \left( \frac{V_0}{g} ight)^2$$L = \frac{mgV_0^3}{2 \sqrt{2} g^2} = \frac{1}{2 \sqrt{2}} \frac{mV_0^3}{g}$

$A$ particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momentum of the particle about $P$ at $t = \frac{V_0}{g}$ is:

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