Projectile hits rod hanging from pivot.

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Homework Help Overview

The problem involves a thin, uniform bar hanging from a pivot that is struck by a small ball. The scenario includes concepts from dynamics and rotational motion, specifically focusing on angular momentum and moment of inertia.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculation of angular momentum before and after the impact, with attempts to derive the moments of inertia for both the ball and the rod. There are questions regarding the correctness of the formulas used and the interpretation of the results.

Discussion Status

Some participants have pointed out potential errors in the application of formulas, particularly regarding the moment of inertia of the rod. There is ongoing clarification about the use of notation in mathematical expressions, and some participants are sharing insights on how to properly format equations.

Contextual Notes

There are indications of confusion regarding the correct syntax for mathematical expressions in the forum, which may affect the clarity of the discussion. Participants are also reflecting on their understanding of the underlying physics concepts.

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.
 
Last edited:
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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
 
BOYLANATOR said:
I obviously used the fraction syntax incorrectly, please let me know what I did wrong
itex gives up if you put non-itex codes like [s u b] inside. Use ^ for sup and _ for sub.
 
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.
 

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