How High Will the Unattached Ball Rise After the Collision?

AI Thread Summary
The discussion revolves around a physics problem involving two 5.00-kg balls and a uniform bar pivoting at its center. The key focus is on applying the conservation of angular momentum to determine how high the unattached ball will rise after a collision with the bar. Participants clarify that while the falling ball does exert a torque, it is not considered an external torque when analyzing the entire system. The approach suggested involves treating the collision similarly to a linear momentum problem but applying it in an angular context. The conversation emphasizes understanding the principles of momentum conservation in rotational dynamics.
boredaxel
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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 \omega
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?
 
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boredaxel said:
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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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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