Analyzing Falling Rod Motion: Angular Velocity and Tip Speed

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SUMMARY

The discussion focuses on the dynamics of a long, thin rod of mass M and length L that falls from a vertical position, pivoting at its lower end. The angular velocity of the rod upon hitting the table is calculated using the moment of inertia formula I = (ML²)/3, resulting in an angular velocity of √(3g/L). To find the speed of the tip of the rod, the angular velocity is multiplied by the length of the rod, confirming that the tip's speed is directly proportional to the angular velocity and the rod's length.

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Homework Statement


A long, thin rod of mass M and length L is standing straight up on a table. Its lower end rotates on a frictionless pivot. A very slight push causes the rod to fall over. As it hits the table, what are a)the angular velocity and b) the speed of the tip of the rod?

I'm having trouble visualizing this problem. If the rod it straight up and the lower end is rotating, isn't the top end also rotating. And I'm not sure how to approach this problem either.

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The Attempt at a Solution

 
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oh ok so i got that first part, finding the angular velocity when it hits the ground
used I= (ML^2)/3 because the axis of rotation is from one end of the rod, and then used energy to find that the angular velocity when it hits the table is square root of (3g/L)

But then for the second part of the question, the velocity of the top part of the rod, I'm not sure how to get. Would it just be the angular velocity I got from the first part multiplied by the length of the rod?
 
roman15 said:
Would it just be the angular velocity I got from the first part multiplied by the length of the rod?
Yep.
 

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