Collision with a pinned and unpinned rod on a frictionless surface

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SUMMARY

The discussion focuses on the dynamics of collisions involving a small object and a long thin rod on a frictionless surface. In the case of a pinned rod, linear momentum is not conserved due to the pin's force, while angular momentum is conserved. For an unpinned rod, both linear and angular momentum are conserved. The energy loss during collisions is determined using the coefficient of restitution, particularly in inelastic collisions where kinetic energy is not conserved. The optimal point of collision for maximum energy loss is debated, with suggestions that hitting the center of the rod minimizes rotational effects.

PREREQUISITES
  • Understanding of linear and angular momentum conservation principles
  • Familiarity with the coefficient of restitution and its application in collision analysis
  • Basic knowledge of kinetic energy equations and conservation of energy
  • Concepts of torque and its relationship to rotational motion
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  • Study the conservation of momentum in both elastic and inelastic collisions
  • Learn how to apply the coefficient of restitution in collision problems
  • Explore the relationship between torque, force, and energy in rotational dynamics
  • Investigate the effects of collision points on energy transfer in rigid body dynamics
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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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