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Vasili
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Homework Statement
A thin, uniform rod of length L and mass M is pivoted from one end. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. Suppose a small metal ball of mass m = 2M is attached to the rod a distance d from the pivot. The rod and the ball are released from rest in the horizontal position. (a) Show that when the rod reaches the vertical position, the speed of its tip is:
[tex]v_{t} = \sqrt{3gL} \sqrt{(1 + 4(\frac{d}{L}))/(1+6(\frac{d}{L}^{2})[/tex]
Homework Equations
The Attempt at a Solution
I've managed to figure out how they got the [tex]\sqrt{3gL}[/tex] bit, but the rest I have no idea. It looks like they're following the [tex]v_{t} = \omega \cdot R[/tex] formula, because [tex]\sqrt{3gL}[/tex] is the angular velocity of the rod as it passes the vertical, but how [tex]\sqrt{(1 + 4(\frac{d}{L}))/(1+6(\frac{d}{L}^{2})[/tex] equals the radius from the pivot to the tip of the rod, I have no idea.
What is the new center of mass of this system when you add the ball to it? The center of mass of just the rod is L/2, so is the center of mass after [tex]\frac{2}{3}(d-\frac{L}{2})[/tex]?
In the formula for the tangential velocity of the tip, shouldn't the radius r just be the length L? Or do you have to use the distance to the center of mass here?
I'll keep working on it. Tips would be appreciated.
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