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Rod collides in a perfect inelastic way with a ball

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Text of the problem: A uniform rod of length $L1$and mass $M$ equal to 0.75 kg is supported by a hinge of negligible mass at one end and is free to rotate in the vertical plane. The rod is released from rest in the position shown. A particle of mass $m$ = 0.50 kg is supported by a thin string of length $L2$ from the hinge. The particle sticks to the rod on contact. What should be the ratio $ \frac {L2}{L1}$ so that $ \theta _{max} = 60°$ after the collision? (Literally copied from Tipler, chapter 10, exercise 67; last edition)

Notice rod goes towards the ball

These are my solutions (based on an answer-pdf)

As we know, angular momentum is conserved due to the fact the torque exerted by the rod on the ball is equal and opposite to the torque exerted by the ball on the rod. Therefore:

$ \Delta L = L_{af.c} - L_{bf.c} $

The system's moment of inertia DOES NOT CHANGE. It is equal to $ \frac{ML_1^2}{3} + mL_2^2 $ What I did wrong was the following : I considered the moment of inertia before the perfectly inelastic collision (PIC) equal to the moment of inertia after the PIC. The moment of inertia before the PIC is equal to the rod one and the moment of inertia after the PIC equals to the one of the system (rod+ball).

$ (\frac{ML_1^2}{3} + mL_2^2) \omega_{af.c} = (\frac{ML_1^2}{3} ) \omega_{bf.c} $

Well, as I do not know neither $ \omega_{bf.c} $ or $ \omega_{af.c} $ we have to find 2 more EQ:

Firstly, in order to obtain $ \omega_{bf.c} $ I analysed the conservation of energy of the rod (as before the collision, the velocity of the system is only due to the rod). Therefore:

$ U_f - U_o + K_f - K_o = 0 $ -----> $ \frac{MgL_1}{2} - MgL_1 + \frac{I \omega_{bf.c}}{2} $

At this point I also had doubts related with the value of initial potential energy. At pdf is pointed out the following: the zero of gravitational potential energy be at a distance $L_1$ below the pivot, so I guessed the value was $MgL_1$ because the center of mass of the rod (initially) it is at a height equal to $L_1$.

Therefore

$$ \omega_{bf.c} = \sqrt{ \frac{3g}{L_1}} $$

Secondly, I calculated the angular velocityof the system after the collision by conservation of angular momentum

$ (\frac{ML_1^2}{3} + mL_2^2) \omega_{af.c} = (\frac{ML_1^2}{3} ) \omega_{bf.c} $

Therefore:

$$ \omega_{af.c} = \frac{ML_1^2}{3(\frac{ML_1^2}{3} + mL_2^2)} \omega_{af.c}$$

Imagine two balls approaching to each other. When both collide stick. $v$ is their initial velocity, therefore KE of the system (ball+ball) is $ KE= 2 \frac{mv^2}{2} $ But final velocity equals to 0. Therefore KE is not conserved (I learned this example from John Rennie). However, energy is conserved before and after the collision due to the lack of friction forces.

Once we have reached this point, we make an energy approach (system's rotational kinetic energy is transformed to gravitational potential energy).

Here I do not obtain the same equation than the answer sheet, which is:

$\frac{-I\omega_{af.c}}{2} + \frac{MgL_{1}}{2}(1-cos \theta_{max}) + mgL_{2}(1-cos \theta_{max}) =0 $

I realized where I got wrong. But I have difficulty determining the height at potential energy. Please could you check if the following reasonings are right?

Taking into consideration the zero of gravitational potential energy is placed horizontally at a distance $L_{1}$ below the pivot

1) Gravitational potential energy of the rod at its initial position equals to $ MgL_1$. Height equals to $L_1$ due to the fact the rod's center of mass is at a distance $L_1$ from the zero of gravitational potential energy.

2) Gravitational potential energy of the rod at its final position (just an instant before hitting the ball) equals to $ \frac{MgL_1}{2}$. The height equals to $ \frac{L_1}{2} $ due to the fact the rod's center of mass is at a distance $ \frac{L_1}{2} $from the zero of gravitational potential energy.

Then my question now is: Why is there not Gravitational potential energy just after the perfectly inelastic collision happens?

http://www.nhvweb.net/nhhs/science/rquinn/files/2013/12/Ch10-Homework-Answers.pdf

Thanks

asked Apr 17 in Physics Problems by JD_PM (490 points)
edited Apr 25 by JD_PM
Note $ \alpha = \frac{m}{M} $  and $ \beta =\frac{L_{2}}{L_{1}} $
The manual provides a very detailed solution. Compare your calculation with that in the manual, line by line, to find out at which step there is a difference.
Yes, that is exactly what I did and obtained  a different result at the line I stood out. But I am sure it will be a silly mistake so don’t bother, I will check it out again
Can I ask you for advise in bibliography? I’m looking for a book about physics challenging problems. I found International Physics Olympiads book and Problems in general physics by Irodov. Any personal suggestion?
No I don't have any personal suggestions. You can easily find problem sets on the internet. You could ask for personal recommendations in the ChatRooms "hbar" or "Problem Solving Strategies" at Physics Stack Exchange.

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