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

AI Thread Summary
The discussion focuses on the dynamics of a thin, uniform rod pivoted at one end with a metal ball attached, analyzing its motion when released from a horizontal position. For part a, the conservation of energy principle is applied, equating initial potential energy to final kinetic energy, leading to the derived speed formula for the rod's tip. The participants express uncertainty about calculating the energy contributions from both the rod and the ball, particularly regarding potential and rotational kinetic energy. Part b addresses finding the center of percussion, where the speed of the rod is equivalent to that when the ball is at the pivot point. The conversation highlights the complexities of rotational dynamics and energy conservation in this system.
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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 = \sqrt{3gL}\sqrt{(1 + 4(d/L))/(1 + 6(d/L)^{2})}

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 Ef=1/2I\omega^{2} + Mg(L/2) 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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