Question 55: Centre of Mass and Rotational Motion

Question

Three equal masses $m$ are kept at vertices $(A, B, C)$ of an equilateral triangle of side a in free space. At $t=0$, they are given an initial velocity $\overrightarrow{V_A}=V_0 \overrightarrow{A C}, \overrightarrow{V_B}=V_0 \overrightarrow{B A}$ and $\overrightarrow{V_C}=V_0 \overrightarrow{C B}$. Here, $\overrightarrow{A C}, \overrightarrow{C B}$ and $\overrightarrow{B A}$ are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is :

Accepted Answer

Here, $$r = \sqrt {{a^2} - {{{a^2}} \over 4}} $$
$$ \Rightarrow r = {{\sqrt 3 a} \over 2}$$
Here, as $${	au_{ext}} = 0$$
we know, $$	au  = {{dL} \over {dt}}$$
$$ \Rightarrow {{dL} \over {dt}} = 0$$ $\Rightarrow$ L = Constant
So, we can apply angular momentum conservation. i.e. angular momentum of the system about any point (Let's take point C) is conserved.
$$ \Rightarrow {L_f} = {L_i}$$
$$ \Rightarrow \overrightarrow {{L_f}}  = o 	imes \overrightarrow {{P_A}}  + \overrightarrow r  	imes m{V_0}\widehat BA + o 	imes \overrightarrow {{P_c}} $$ (as $${P_B} = m{v_0}\widehat {BA}$$ and $$\overrightarrow L  = \overrightarrow r  	imes \overrightarrow P $$)
$$ \Rightarrow \overrightarrow {{L_f}}  = \left( {{{\sqrt 3 a} \over 2}} ight)m{v_0}\sin 90^\circ $$ (As $$\left| {\overrightarrow r } ight| = {{\sqrt 3 a} \over 2}$$)
$$ \Rightarrow \overrightarrow {{L_f}}  = {{\sqrt 3 } \over 2}am{v_0}$$ (outwards & $$ \bot $$ to plane)
$$ \Rightarrow {L_f} = {{\sqrt 3 } \over 2}am{v_0}$$
Hence, option D is correct.

Answer

$\frac{1}{2}$ a $\mathrm{mV}_0$

Answer

3 a $\mathrm{mV}_0$

Answer

$\frac{3}{2}$ a $\mathrm{mV}_0$

Answer

$\frac{\sqrt{3}}{2}$ a $\mathrm{m}V_0$

Centre of Mass and Rotational Motion

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