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Consider a uniform rod of massM=4m and length l pivoted about its centre. A mass m moving with velocity v making angle \theta =\frac{\pi }{4} to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod - mass system just after the collision is :   
Option: 1 \frac{4}{7}\frac{v}{l}

Option: 2 \frac{3\sqrt{2}}{7}\frac{v}{l}

Option: 3 \frac{3}{7\sqrt{2}}\frac{v}{l}  

Option: 4 \frac{3}{7}\frac{v}{l}

Answers (1)

best_answer

 

Let us conserve angular momentum about O:-

           So, L_i=\left ( \frac{mv}{\sqrt2} \right )\times \frac{l}{2}, where \left ( \frac{mv}{\sqrt2} \right ) is linear momentum and \left ( \frac{l}{2} \right ) is the distance from centre O.

Now, L_f=I\omega

Here, I=\frac{4ml^2}{12}+{m\left (\frac{l}{2} \right )^2}=\frac{7ml^2}{12}

So, L_i=L_f\Rightarrow \omega=\frac{6v}{7\sqrt2 l}= \frac{3\sqrt2v}{7 l}

So the correct graph is given in option 2.

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vishal kumar

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