# Tension in the rod

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On a smooth table two particles of mass m each, travelling with a velocity v$_0$
in opposite directions, strike the ends of a rigid massless rod of length l, kept
perpendicular to their velocity. The particles stick to the rod after the collision.Find
the tension in rod during subsequent motion.

I tried to conserve momentum ,
2mv$_0$R = I$\omega$
2mv$_0$R = (2mR$^2$ ) $\omega$

got the angular velocity of combined motion a ($v_0$)/(R)
But how to proceed.

edited Dec 23, 2016
But there are two masses , so will we multiply it by 2 as you said earlier' centripetal force for each mass.'
No, the tension is only what is required for 1 mass. If there was only 1 mass and the rod is attached to a pivot, the tension will be the same.
Could you explain it more briefly
Or could you post an answer
Consider each mass separately. The only force on each is the tension in the rod. This force is centripetal, so it must equal $mv^2/R$. The tension is the same throughout the rod. It does not double just because there is a mass on both ends.

We know centripetal force = m$\omega$R$^2$
=mv$^2$/R
Hence the force is 2mv$^2$/L