- #1
n00bot
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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
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