Collision with a pinned and unpinned rod on a frictionless surface

AI Thread Summary
In the discussion about collisions involving a pinned and unpinned rod on a frictionless surface, it is established that linear momentum is not conserved when the rod is pinned, while angular momentum is conserved in both scenarios. For the pinned rod, energy loss occurs during the collision, while in the unpinned case, both linear and angular momentum are conserved, and there is no energy loss. The coefficient of restitution can be used to calculate energy loss in the pinned scenario, while the collision with the unpinned rod is completely inelastic. The optimal point for energy loss in the collision is debated, with suggestions that hitting the center of the rod may minimize rotation and energy loss. Overall, the complexities of conservation laws and energy transfer in these collision scenarios are highlighted.
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1. The Problem: (Collision with a thin rod -- pinned and unpinned)
A small object with mass n and an initial velocity of v sticks to a long thin rod of mass m and length l. The motion takes place on a horizontal frictionless surface. Answer the questions for the situation where
a) the rod is pinned at its center (but able to spin frictionlessly), the object rebounds, and the coefficient of restitution is r.
b) The rod is not pinned and the object sticks to the rod.

2. The Questions:
1. In each situation, is linear or angular momentum conserved? Explain.
2. Determine the energy lost in the collision.
3. Where should the small object collide with the rod to maximize the loss of energy in the collision? Explain.


3. The Attempt at a Solution :
1. a) The pin applies force on the rod, so linear momentum is not conserved within the rod-object system (the pin transfers some momentum to the surface). Angular momentum is conserved since the rod is free to rotate.
b) Both angular and linear momentum are conserved: the rod is free to rotate, and free to move. Rotation and translation.

2. a) Elastic collision, so KEinitial = KEfinal. I assume energy would be transferred to the rod from the object, but the system is constant. Maybe?

b) Since it is a completely inelastic collision, I know that KEinitial > KEfinal. My best guess to determine loss of energy would be to set up a conservation of energy equation, and have mobj-initialvobj-initial2 + 0 = (mobj + mrod)vfinal2. I'd then solve for vfinal, and plug Kinitial=mobj-initialvobj-initial2
Kfinal= (mobj + mrod)vfinal Is that correct?


3. a) if KEinitial = KEfinal, then no energy would be lost, and it wouldn't make a difference where the rod was hit. But... what about the pin? (How could that be right?!)

b)I assume the max loss of energy would occur with a collision at the center, because that would cause the bar to rotate the least (or, not at all). I can't really figure out where to go with this, though, since there's no friction (so, where would the energy go?) How do I go about figuring this out?

Thanks
 
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hehe, I am new here, so i hope we all can work together.

well i think in

1. a)

linear and angular is not conserved.
reason: energy loss during collision (sound, heat, etc) and total initial energy is not equal to total final energy

1. b)

linear and angular in conserved, since is inelastic collision with no friction on surface

2. a) use the coeficient of restitution formula <br /> \textrm{C.O.R.} = <br /> \frac{|\vec{v_2} - \vec{v_1}|}<br /> {|\vec{u_2} - \vec{u_1}|} <br />

substitute v1 and v2 in kinetic equation... the differences will get the energy loss.

2. b) there is no energy loss.

3. a) the edge of the rod, hmm, i think we can explain by relationship using formula of torque, force, and energy i guess

3. b) anywhere is the same, i think xP
anyway, please help me if i was wrong somewhere and I'm sorry if I'm wrong.. and i also sorry because i don't know whether i can leave my answer or not.. because i think only tutor can post answer here.. notify me if i did something wrong..
thanks
 
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