A particle is projected at time $t=0$ from a point $O$ with a speed $v_{0}$ at an angle of $45^{\circ}$ to the horizontal. Find the magnitude and the direction of the angular momentum of the particle about the point $O$ at time $t=v_{0} / g$.

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Aicharya · Accepted Answer

Let us take the origin at $\mathrm{P}, \mathrm{X}$-axis along the horizontal and Y -axis along the vertically upwards direction as shown in figure. For the horizontal motion during the time O to t, $\begin{aligned} & v_x=v_{0} \cos 45^{\circ}=\frac{v_0}{\sqrt{2}} \ & \text{and } x=v_x t=\frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{g}=\frac{v_0^2}{\sqrt{2} g} \end{aligned}$ For vertical motion, $\begin{aligned} & v_{Y}=v_{0} \sin 45^{\circ}-gt=\frac{v_{0}}{\sqrt{2}}-g=\frac{1-\sqrt{2}}{\sqrt{2}} v_{0} \ & \text{and } y=\left(v_{0} \sin 45^{\circ}\right) t-\frac{1}{2} g t^2 \ & y=\frac{v_0^2}{\sqrt{2} g}-\frac{v_0^2}{2 g}=\frac{v_0^2}{2 g}(\sqrt{2}-1) \end{aligned}$ The angular momentum of the particle at time t about the origin is $\begin{aligned} & L=\vec{r} \times \vec{p}=m \vec{r} \times \vec{v} \ & =m(\vec{i} x+\vec{j} y) \times\left(\vec{i} v_x+\vec{j} v_y\right) \ & =m\left(\vec{k} v_y-\vec{k} y v_x\right) \ & =m \vec{k}\left[\left(\frac{v_0^2}{\sqrt{2} g}\right) \frac{v_0}{\sqrt{2}}(1-\sqrt{2})-\frac{v_0^2}{2 g}(\sqrt{2}-1) \frac{v_0}{\sqrt{2}}\right] \ & =-\vec{k} \frac{m v_0^3}{2 \sqrt{2} g} \end{aligned}$ Thus, the angular momentum of the particle is $\frac{m_0^3}{2 \sqrt{2} g}$ in the negative Z -direction, i.e., perpendicular to the plane of motion, going into the plane.

A particle is projected at time \(t=0\) from a point \(O\) with a speed \(v_{0}\) at an angle of…

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