Rank Pendulums Based on Max Speed

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SUMMARY

The discussion focuses on ranking pendulums based on their maximum speed, utilizing the principle of conservation of mechanical energy. The key equation derived is v = √(2gΔh), where v represents maximum speed, g is the gravitational acceleration, and Δh is the height of release. It is established that the mass of the pendulum does not affect the maximum speed, as it cancels out in the energy equations. Participants confirm that the maximum speed depends solely on the height from which the pendulum is released and the local gravitational acceleration.

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Homework Statement



Rank each pendulum on the basis of its maximum speed.
http://img148.imageshack.us/img148/9674/picture3op6.png

Homework Equations



m*g*\Delta = 1/2*m*v^{2}



The Attempt at a Solution




At first I thought that all the Max speeds were the same, but I was wrong.

I know that the kinetic energy is equal to the change in potential energy since I solve the first 2 parts correctly.


If I follow the equation, I know the mass will cancel out. So am I solving for v^{2}?


So if I simplify the equation is it:

v = \sqrt{2*g*\Deltah}

Then would i just apply it to all the situations?

Thanks

Sorry for the bad format for the equations.
 
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Duckboy said:

Homework Equations



m*g*\Delta = 1/2*m*v^{2}

Yes, you have the important part right here. We assume that the mechanical energy, E = K + U (kinetic+potential energy) is conserved in the pendulum swing, because gravity is a conservative force and we ignore air resistance and friction in the bearing the pendulum swings on.

So when the pendulum is initially released, all of the mechanical energy is in gravitational potential energy (relative to the bottom of the swing), all of which is converted into kinetic energy when the pendulum reaches the bottom. This let's you write the equation above.

You solved this for v^2 = 2gh correctly. So this tells you that in an entirely gravitational process of this sort, the mass is irrelevant and that the speed at the bottom of the swing will depend only on the local gravitational acceleration and the height of release of the pendulum bob. (Thus, in sorting your examples, some will be of equal rank!)
 
Thanks! I just needed to be sure.
 

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