What Is the Speed of the Rod's Tip When Released from a Horizontal Position?

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SUMMARY

The discussion focuses on the dynamics of a thin, uniform rod with moment of inertia I = 1/3ML², pivoted at one end, and a metal ball of mass m = 2M attached at a distance d from the pivot. When released from a horizontal position, the speed of the rod's tip at the vertical position is derived as v = √(3gL)√((1 + 4(d/L))/(1 + 6(d/L)²). Additionally, the discussion addresses finding the value of d/L where the speed of the rod matches that of d = 0, identifying this point as the center of percussion or "sweet spot" of the rod.

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A thin, unirform rod (I=1/3ML^2) of length L and mass M is pivoted about one end. A small metal ball of mass m=2M is attached to the road a distance d from the pivot. The rod and ball are realeased from rest in a horizontal position and allowed to swing downward without friction or air resistance.

a) show that when the rod reaches the vertical position, the speed of its tip is:
v = [tex]\sqrt{3gL}\sqrt{(1 + 4(d/L))/(1 + 6(d/L)^{2})}[/tex]

b) At what finite value of d/L is the speed of the rod the same as it is for d=0? (This value of d/L is the center of percussion, or "sweet spot" of the rod.)

For part a i know its Ei = Ef, but I am not sure what you should be using to calculate it..
The rod has Ei = MgL, and [tex]Ef=1/2I\omega^{2} + Mg(L/2)[/tex] But i don't know what i should calculate for the ball, potential at the top and rotational kinetic at the bottom?

i haven't even looked at part b yet.
 
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