Projectile hits rod hanging from pivot.

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BOYLANATOR
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Homework Statement


A thin, uniform bar, 2, long and weighing 90N is hanging vertically from the ceiling by a frictionless pivot. It is struck by a small 3kg ball, 1.5m below the ceiling, initially traveling horizontally at 10 m/s. The ball rebounds in the opposite direction with a speed of 6 m/s.

Homework Equations



Lbefore = Lafter

L = Iω

Irod = [itex]\frac{1}{2}[/itex]MR2

Ipoint = MR2

The Attempt at a Solution



At the point of impact the ball can be thought of as a particle in circular motion about the pivot with radius 1.5m and tangential velocity 10m/s.
The change in velocity is 16m/s. So the effective change in the angular momentum of the ball is

ΔLball = Iball Δωball = [itex]\frac{I<sub>ball</sub>Δv}{R<sub>ball</sub>}[/itex]

This is equal to the change in the angular momentum of the rod (opposite direction):

ΔLrod = Irod Δωrod = [itex]\frac{I<sub>ball</sub>Δv}{R<sub>ball</sub>}[/itex]

Insert values for I and rearrange :

Δωrod = (2*mball*rball*Δvball) / (mrod*rball2)

This gives ω = 3.92 rad/s , the given answer is 5.88 rad/s.
 
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I obviously used the fraction syntax incorrectly, please let me know what I did wrong. Thanks
 
BOYLANATOR said:

Homework Statement


A thin, uniform bar, 2, long and weighing 90N is hanging vertically from the ceiling by a frictionless pivot. It is struck by a small 3kg ball, 1.5m below the ceiling, initially traveling horizontally at 10 m/s. The ball rebounds in the opposite direction with a speed of 6 m/s.


Homework Equations



Lbefore = Lafter

L = Iω

Irod = [itex]\frac{1}{2}[/itex]MR2

The formula for the moment of inertia of the rod is not correct.


ehild
 
As haruspex said, _ for sub, but enclose subscript between curly thingies {} :smile:

[itex]ΔL_{ball} = I_{ball} Δω_{ball} = \frac{I_{ball}Δv}{R_{ball}}[/itex]

written as

ΔL_{ball} = I_{ball} Δω_{ball} = \frac{I_{ball}Δv}{R_{ball}}

ehild
 
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Ah yes. I should stick to deriving the moments of inertia, my memory doesn't serve me well. Thanks.