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Pivoting Rod and Ball - Rotational Dynamics

  • Thread starter DelPopolo
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  • #1
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Homework Statement


A cylindrical rod 26.0 cm long with a mass of 1.20 kg and a radius of 1.50 cm has a ball of diameter of 7.80 cm and a mass of 2.00 kg attached to one end. The arrangement is originally vertical and stationary, with the ball at the top. The apparatus is free to pivot about the bottom end of the rod.

(a) After it falls through 90°, what is its rotational kinetic energy?


Homework Equations


KE=1/2I[tex]omega[/tex]^2
I=(2mr^2)/5
I=(2mL^2)/3

The Attempt at a Solution



Moment of inertia=I of rod + I of ball?
I=(2*m*(r+L)^2)/5)+(2mL^2)/3
I'm lost as to finding [tex]\omega[/tex] so I can find KE.

Please point me in the right direction.
Thank you
 
Last edited:

Answers and Replies

  • #2
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Moment of Inertia of a rod about pivot is

[tex]I_r=\frac{mL^2}{3}[/tex]

And if you take level of pivet as zero level,after rod falls through 90°,total potential energy will be zero

so we have

[tex]PE_{initial}=\frac{1}{2}I_{total}\omega^2[/tex]
 
  • #3
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I=(2*m*(r+L)^2)/5)+(mL^2)/3
I=.0985

so

KE=1/2(.0985)[tex]\omega[/tex]^2

I need to find omega so I can find kinetic E. Correct?

How does one find omega? Obviously gravity affects the speed but gravity isnt the same over the entire rotation so how does that work out?
 
Last edited:
  • #4
190
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What's the potential energy of this system,when it's at rest?
 

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