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A uniform rod of mass '1 kg' and length '\sqrt{2} m' is lying on a horizontal table. Its one end is connected with a massless inextensible string of length 'l m' which is connected with a hinge as shown in the figure.

A particle of mass '1 kg' travelling with a speed of '2 m/s' hits the rod perpendicularly and sticks to it. Find the angular velocity of the rod just after the impact.

 

 

Option: 1

\sqrt{2} rad/s


Option: 2

2\sqrt{2} rad/s


Option: 3

\sqrt{3} rad/s


Option: 4

3\sqrt{3} rad/s


Answers (1)

best_answer

Conserving angular momentum about point A,

mv_{0}L = 2m (\frac{3l}{4}\omega - v\sin \theta) + I_{CM}\omega     ----- 1

I_{CM} = \frac{ml^2}{12} + \frac{ml^2}{16} + \frac{ml^2}{16}

I_{CM} = \frac{5}{24} ml^2

Applying conservation of their momentum in the direction perpendicular to the string:

\frac{mv_{0}\sin \theta}{2} = \frac{3L}{4} \omega \sin \theta - v                ----- 2

Solving equations 1 and 2:

\omega = \frac{v_{0}}{l} = \frac{2}{\sqrt{2}} = \sqrt{2} rad/s

v = \frac{v_{0}}{3\sqrt{2}} = \frac{2}{3\sqrt{2}} = \frac{\sqrt{2}}{3} m/s

Posted by

vishal kumar

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