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Energy equivalence between linear and rotational motion

  • Thread starter Waxbear
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  • #1
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Technically this is a homework question because it's from an assignment I'm doing as practice for my exam tomorrow.

Imagine a rod standing on a table, the base of the rod is attached to the table with a hinge, so that the rod is able to swing between standing position and parallel with the table. The rod has the length L.

The rod is standing upright, so that it's in equilibrium. A ball comes flying through the air, and hits the rod at 1/2 L, after which the ball drops vertically to the table. Consider the transfer of kinetic energy to be instant.

The equivalence between kinetic energy in the ball and in the rotational motion of the rod should be: [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]=[tex]\frac{1}{2}[/tex]I[tex]\omega[/tex][tex]^{2}[/tex] right?

Now consider the ball having the same mass and the same speed as before, but now it hits the rod at L. wouldn't the initial rotational velocity of the rod be larger? if so, why? the kinetic energy in the ball hasn't changed.

It's probably something trivial and stupid that I've overlooked. So if anyone would care to explain what I've missed, it would be much appreciated :-)

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
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whoops, sorry for messing up with the template
 
  • #3
tiny-tim
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Welcome to PF!

Hi Waxbear! Welcome to PF! :smile:

(have an omega: ω and try using the X2 icon just above the Reply box :wink:)

use conservation of angular momentum not energy

momentum and angular momentum are always conserved in collisions, energy usually isn't :smile:
 
  • #4
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Thanks Tim, awesome forum by the way ;)

I checked the answer to the assignment, and they did use angular momentum/momentum conservation as you said. But the fact that energy isn't conserved here bothers me, since i can't see why it isn't!
If we say that no kinetic energy is converted to heat, and since there is no linear motion after the collision (except for the ball dropping, which is just the potential energy being converted) then where does the rest of the energy go?
 
  • #5
D H
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But the fact that energy isn't conserved here bothers me, since i can't see why it isn't!
If we say that no kinetic energy is converted to heat, and since there is no linear motion after the collision (except for the ball dropping, which is just the potential energy being converted) then where does the rest of the energy go?
In most collisions a lot of the energy is converted to heat, and if not to heat, to broken and bent parts. Energy is conserved in collisions, just not in any useful way.
 
  • #6
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I can imagine that moving the collision point closer to the base of the rod would cause the rod to vibrate or "wobble" more, thus giving it a slower angular velocity, since some energy is lost to this oscillating motion. I was just curious whether or not this loss of kinetic energy was purely due to real-world factors like this, or if there was a dependency between collision height and initial angular speed, even in an idealized model with a totally rigid rod. In other words, whether or not this dependency is solely due to factors unaccounted for, or if there is a theoretical dependency.

But from your answer i take it that this is only a matter of factors present in the "non-ideal" system.
 
Last edited:
  • #7
tiny-tim
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Hi Waxbear! :wink:
… the fact that energy isn't conserved here bothers me, since i can't see why it isn't!
If there'd been a smooth curve, energy would have been conserved.

The jerk makes all the difference. :smile:

Rule of thumb … if there's a bang, then energy is not conserved! :biggrin:
 
  • #8
42
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okay, thanks for clearing that up :)
 

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