Question 362: Centre of Mass and Rotational Motion

Question

A small particle of mass $$m$$ is projected at an angle $$	heta $$ with the $$x$$-axis with an initial velocity $${v_0}$$ in the $$x$$-$$y$$ plane as shown in the figure. At a time $$t < {{{v_0}\sin 	heta } \over g},$$  the angular momentum of the particle is ................,

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

Accepted Answer

Position vector of the particle from the original at any time t is
$$\overrightarrow r  = {v_0}t\cos 	heta \widehat i + \left( {{v_0}t\sin 	heta  - {1 \over 2}g{t^2}} ight)\widehat j$$
As Velocity vector, $$\overrightarrow v  = {{d\overrightarrow r } \over {dt}}$$
$$	herefore$$ $$\overrightarrow v  = {d \over {dt}}\left( {{v_0}t\cos 	heta \widehat i + \left( {{v_0}t\sin 	heta  - {1 \over 2}g{t^2}} ight)\widehat j} ight)$$
$$ = {v_0}\cos 	heta \widehat i + \left( {{v_0}\sin 	heta  - gt} ight)\widehat j$$
Angular momentum of the particle about the origin is
$$\overrightarrow L  = m\left( {\overrightarrow r  	imes \overrightarrow v } ight)$$
$$\eqalign{
  & \overrightarrow L  = m\left[ {{v_0}\,\cos 	heta t\widehat i + \left( {{v_0}\,\sin 	heta t - {1 \over 2}g{t^2}} ight)\widehat j} ight]  \cr 
  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, 	imes \left[ {{v_0}\,\cos \,	heta \widehat i + \left( {{v_0}\,\sin 	heta  - gt} ight)\widehat j} ight] \cr} $$
$$ = m{v_0}\cos 	heta t\left[ { - {1 \over 2}gt} ight]\widehat k$$
$$ =  - {1 \over 2}mg{v_0}{t^2}\cos 	heta \widehat k$$

Answer

$$ - mg\,{v_0}{t^2}\cos 	heta \widehat j$$

Answer

$$mg\,{v_0}t\cos 	heta \widehat k$$

Answer

$$ - {1 \over 2}mg\,{v_0}{t^2}\cos \,	heta \widehat k$$

Answer

$${1 \over 2}mg\,{v_0}{t^2}\cos 	heta \widehat i$$

Centre of Mass and Rotational Motion

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