Why is my calculation for the speed of a falling rod off by sqrt(2)?

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The discussion centers on calculating the speed of a falling rod pivoted at one end, with the correct final speed being sqrt(3gL). The initial poster's calculations were incorrect by a factor of sqrt(2), leading to confusion about the energy transformation from gravitational potential energy to rotational kinetic energy. Key points include using the height of L/2 for the center of mass and the moment of inertia for a rod rotating about its end, which is 1/3mL^2. The thread also includes a reminder about posting guidelines on the forum. Understanding the energy conversion and correct formulas is crucial for accurate calculations.
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A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
My answer is off by sqrt(2) from the correct answer, sqrt(3gL), and I don't understand why. Please help!
 
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the gpe is being transformed into KE rotational. you would use the height as L/2 since the center of mass is changing that much distance.
the moment of inertia of a stick rotating about the end is 1/3mL^2

mg(0.5L) = 0.5I\omega^2
 
faculaganymede said:
A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
My answer is off by sqrt(2) from the correct answer, sqrt(3gL), and I don't understand why. Please help!

You should not have posted this here. Did you not read https://www.physicsforums.com/showthread.php?t=89899 thread ?


regards
marlon
 
sorry, that was my first post. didn't know better.
 
Thanks Andrew!
 

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