Angular momentum ball problem

  • Thread starter boredaxel
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  • #1
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Homework Statement


A 5.00-kg ball is dropped from a height of 12.0 m above
one end of a uniform bar that pivots at its center. The bar has mass
8.00 kg and is 4.00 m in length. At the other end of the bar sits
another 5.00-kg ball, unattached to the bar. The dropped ball sticks
to the bar after the collision. How high will the other ball go after
the collision?



Homework Equations


L= r X p
L = I [tex]\omega[/tex]
mgh =1/2 mv^2

The Attempt at a Solution


The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
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Welcome to PF!

A 5.00-kg ball is dropped from a height of 12.0 m above one end of a uniform bar that pivots at its center.
The bar has mass 8.00 kg and is 4.00 m in length. At the other end of the bar sits another 5.00-kg ball, unattached to the bar.
The dropped ball sticks to the bar after the collision.
How high will the other ball go after the collision?

The solution seems to require the conservation of angular momentum. But i am not sure how angular momentum could be conserved? Doesnt the weight of the falling ball provide a net external torque to the system?

Hi boredaxel! Welcome to PF! :smile:

Yes, it does provide a torque, but that doesn't matter, because it isn't an external torque …

at least, if you consider the whole system together, it isn't external! :wink:

Angular momentum, like momentum, is always conserved.

Hint: treat this exactly the same way as you would if the ball was hitting a block with another ball directly the other side …

you'd use conservation of (linear) momentum, and v1f = v2f, wouldn't you?

Well, do the same, except … "angularly"! :biggrin:
 
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