Finding final kinetic energy of a system

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SUMMARY

The discussion centers on calculating the final kinetic energy of a system involving a uniform solid sphere (mass M, radius R) with a bullet (mass m) fired into it at a distance of 2R/3 from the center. The key principle applied is the conservation of angular momentum, as linear momentum is not conserved in this scenario. The placement of the bullet significantly impacts the final kinetic energy, confirming that the distance from the pivot point (the spike) is crucial for accurate calculations. The problem emphasizes the importance of understanding the rotational dynamics of the system.

PREREQUISITES
  • Understanding of angular momentum conservation
  • Familiarity with rotational dynamics of rigid bodies
  • Knowledge of kinetic energy calculations in rotational systems
  • Basic physics concepts related to collisions and energy transfer
NEXT STEPS
  • Study the principles of angular momentum in detail
  • Learn about the moment of inertia for different shapes, particularly spheres
  • Explore the effects of distance from the pivot point on rotational motion
  • Investigate energy conservation in inelastic collisions
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Physics students, educators, and anyone interested in understanding the dynamics of rotating systems and energy conservation principles in mechanics.

stripes
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Homework Statement



A uniform solid sphere of mass M and radius R has a light, thin metal spike driven through it, so it is free to turn about this spike. So imagine a bowling ball on a stick that is free to rotate about this stick.

A bullet of mass m is fired at speed v into the ball at a distance 2R/3 from the center, and it comes to rest in the sphere. What is the final kinetic energy? Hint: angular momentum.

I recall my instructor saying it might or might not matter where the bullet ends up...but we never really concluded anything. If it does matter, let's say it ends up half way through the ball.

Homework Equations



The Attempt at a Solution



So we use angular momentum because it is conserved in this case, while linear momentum is not.

To be honest I'm not sure where to begin...I'm trying to put things together but I'm not sure where the distance 2R/3 comes into play, for example. If someone could help me out, I would really appreciate it.
 
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A likely scenario is that the bullet strikes the bowling ball horizontally at a height of 2/3 R above the horizontal center line of the ball:

attachment.php?attachmentid=63778&stc=1&d=1383948205.gif
 

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hi stripes! :smile:
stripes said:
A uniform solid sphere of mass M and radius R has a light, thin metal spike driven through it, so it is free to turn about this spike. So imagine a bowling ball on a stick that is free to rotate about this stick.

A bullet of mass m is fired at speed v into the ball at a distance 2R/3 from the center, and it comes to rest in the sphere. What is the final kinetic energy?

i think this question is totally incomprehensible :redface:

does it mean 2/3R from the stick ?

and yes of course it matters how far in the bullet stops!
 
Hey guys. I have since solved the problem. Also, I copied it verbatim as per the assignment. If you ever check my other homework questions in physics, you will find that it's not uncommon for this instructor to word them poorly, like this. I just went to see him and clarify. Thanks anyways.
 
gneill said:
A likely scenario is that the bullet strikes the bowling ball horizontally at a height of 2/3 R above the horizontal center line of the ball:

attachment.php?attachmentid=63778&stc=1&d=1383948205.gif
It will have to be a vertical axis, or there will be no 'final' KE; it will oscillate.
 
haruspex said:
It will have to be a vertical axis, or there will be no 'final' KE; it will oscillate.

Ah. True. So the "spike" is vertical and the diagram I drew is looking down on the scene from above.
 

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