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A small particle of mass m is projected at an angle $\Theta$ with the x-axis with an initial velocity $\upsilon _{0}$ in the x-y plane as shown in the figure. At a time $t< \frac{\upsilon _{0}\sin \Theta }{g},$ the angular momentum of the particle is

where $\hat{i},\hat{j}\, and\, \hat{k}$ are unit vectors along x,y and z-axis respectively.

Answers (1)

$\vec{v}_x = \left ({v_{0}} \right \cos \Theta)\hat{i}$

$\vec{v}_y = \left ( v_{0} \right\sin \Theta -gt )\hat{j}$

$\vec{v} ={v}_x\hat{i}\:+{v}_{y}\hat{j}$

$\vec{v} = v_{o}\cos \Theta \hat{i} + \left (v _{0} sin \Theta \right - gt )\hat{j}$

$\vec{r} = v_{o} \cos \Theta \hat{i} + \left (v_{o} \sin \Theta t- {\tfrac{1}{2}} gt^2 \right )\hat{j}$

Angular momentum

$= \vec{L} = m\left ( \vec{r} \times \right\vec{v} )$

$L = m (v_0cos\Theta \hat{i}+(v_0sin\Theta t-\frac{1}{2}gt^2)\vec{j})\times (v_0cos\Theta \hat{i}+(v_0sin\Theta-gt))\hat{j}$

$L = m [v^2_0cos\Theta sin\Theta t - v_0gt^2cos\Theta]\hat{k}+(v^2_0sin\Theta cos\Theta t - \frac{1}{2}gt^2v_0cos\Theta (-\hat{k})$

$L =m (v^2_0sin\Theta cos\Theta t \hat{k} - v_0gt^2 cos\Theta (\hat{k}) - v^2_0sin\Theta cos\Theta t \hat{k} +\frac{1}{2}mgv_0t^2cos\Theta \hat{k}$

$L =m[- \frac{1}{2}v_0gt^2cos\Theta \hat{k}]$

$L =\frac{1}{2}mgv_0t^2cos\Theta \hat{k}$

Posted by

Safeer PP

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