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Rotation of L shaped rods after impulse

1 vote

Two identical rod each of mass $m$ and length $\ell$ joined at one end and
free to rotate about this end kept on smooth horizontal surface. An
horizontal impulse $J$ is given to rod AB at centre and perpendicular to
rod AB.

I am just able to deal the angular velocity about centre of but unable to proceed for individual rods.

Unable to proceed for second part.

asked Dec 7, 2018 in Physics Problems by koolman (4,286 points)
edited Dec 8, 2018 by sammy gerbil

1 Answer

3 votes
Best answer

The hardest part of these questions is working out what they mean! Here it is not clear (i) whether joint B is free to translate or is fixed to the ground, and (ii) whether the rods are joined rigidly at B (so that they remain at $90^{\circ}$ at all times), or pinned loosely so the angle between them can change freely.

If the rods are joined rigidly at B then there will be a torque exerted on AB by rod BC as well as an impulsive force, and vice versa for AB acting on BC. This is because for two 2D objects to be attached rigidly together, and maintain the same relative positions, they must be pinned at 2 distinct points. The diagram suggests there is only one pin at joint B. Also, Part 2 of the question does not mention any torque. So I think the rods must be pinned loosely at B by a single pin. If this pin is also fixed to the ground then the impulse on AB will have no effect on BC because the reaction at the pin will be transmitted to the Earth, which is so massive that it does not move.

So I think joint B must be pinned loosely with a single pin and is free to translate. This makes sense because rods AB and BC can then have different angular velocities (see options in Part 1).

Your attempt assumes the two rods are rigidly connected, which is a reasonable interpretation of the question, but I think for the above reasons it must be wrong.

You are correct about the position of the COM of the 2 rods. But you have made an error in applying the Parallel Axis Theorem. This always starts with an axis through the COM and gives you the MI about an axis parallel to that through the COM. You have started at end B of the rod, not at the COM, so your calculation is incorrect. You should get $\frac{5}{12}m\ell^2$.


Rod AB is struck at its COM. If it were not pinned to rod BC it would move upwards on the page, without rotating. This tells us that the reaction impulse $K$ at the pin is purely downwards on the page when acting on AB, and purely upwards when acting on BC.

Impulse $K$ acts through the COM of rod BC so this rod does not rotate : $\omega_2=0$. It moves upwards with velocity $v_2$ such that $K=mv_2$.

The impulses on AB can be resolved into a linear impulse $J-K$ acting upwards at the COM and a clockwise couple $K\frac{\ell}{2}$ about the COM. See Is net torque is not zero about all points on the rod for a linearly accelerating rod? The COM of rod AB moves upwards with velocity $v_1$ such that $J-K=mv_1$. It also rotates with angular velocity $\omega_1=\frac{v_1-v_2}{ \ell/2}$ such that $K\frac{\ell}{2}=I_1\omega_1$ where $I_1=\frac{1}{12}m\ell^2$ is the moment of inertia of AB about its COM.

We now have enough equations so that we can find the following unknowns : $$\omega_1=\frac{6J}{5m\ell}, \omega_2=0$$ $$v_1=\frac{4J}{5m}, v_2=\frac{J}{5m}$$ $$K=\frac15 J$$

In Part 1 options A, B, C are correct, and in Part 2 option A is correct.

answered Dec 8, 2018 by sammy gerbil (28,806 points)
edited Dec 13, 2018 by sammy gerbil