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The problem is like this: two blocks(considered to be point masses) of masses $m$ are simultaneously moving with uniform velocities $v_0$ towards the rods and they hit the edges of rod and stick to it(i.e, perfectly inelastic collision). Each of the rods have a mass $m$(same as the blocks) and have length $l$. They are hinged together in the middle by a friction less hinge. Find the velocity of the hinge just after collision and the angular velocities of the rods about the hinge.

My approach:

Let the vel. of the hinge after collision be $v$. Then by conservation of linear momentum, $$mv_0 + mv_0 = 4mv \Rightarrow v= v_0/2$$ Note: I believe that the centre of mass of the system(two rods and the blocks sticked to them) is at the hinge just after collision. enter image description here

Let the angular velocity of the rods about the hinge be $\dot\theta$ Now, let us conserve the angular momentum of the system abot point $P$. $$2lmv_0=0+2l \cdot m \cdot(v_0/2+\dot\theta l)+ \frac{3l}{2} \cdot m \cdot(v_0/2+\dot\theta \frac{l}{2}) + \frac{ml^2}{12} \cdot \dot{\theta} + \frac{l}{2} \cdot m \cdot(v_0/2+\dot\theta \frac{l}{2}) - \frac{ml^2}{12} \cdot \dot{\theta}$$

On solving, we get $\dot{\theta}=0$

So, according to my equations, the angular velocity is zero. Are my calculations oK? Or am I missing something?


1 Answer 1


Qualitatively this is what I see happening.

The total angular momentum of the rods before the collisions was zero as there was no rotation of the two rods as a whole, or of either rod individually about the hinge. Neither do the rods have any initial linear momentum.

Each of the individual masses has a torque about the hinge. But they are equal in magnitude and opposite in direction for a net angular momentum of zero. But each has linear momentum.

So the total initial angular momentum of the system (two masses plus rod) is zero and the total linear momentum is the sum of the two masses. Since no external forces act upon the system during the collisions the total linear and angular momentum following the collision must also be the same as before the collisions for conservation of angular momentum.

After the collision the two masses stick to the rods causing each rod to rotate about the hinge. The two masses still have the same equal and opposite angular momentum for a total of zero after the collision. Each rod also acquired an angular momentum after the collision, but they are equal in magnitude and opposite in direction for a net angular momentum of zero.

The total net angular momentum of the system before and after the collision is zero and angular momentum is conserved.

The total initial linear momentum of the two masses equals the total linear momentum of the center of mass (hinge) of the system of the two masses and rods after the collision and therefore linear momentum is also conserved.

Bottom line in response to title of your post, I believe each rod will individually rotate about the hinge but with equal and opposite angular momentum for a net angular momentum of zero.

Hope this helps.


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