Finding the Velocity of the Bob on a Pendulum

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    Pendulum Velocity
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To find the speed of the bob at the bottom of a 2.0 m pendulum released from a 25-degree angle, the potential energy (PE) at the starting height is converted into kinetic energy (KE) at the lowest point. The correct height calculation involves using the difference between the initial height and the lowest point, which is 2 - 2cos(25). Initially, the user calculated the velocity as 0.59 m/s but later revised it to 1.92 m/s after clarifying the height calculation. The discussion emphasizes the importance of accurately determining height in energy conservation problems. Understanding the relationship between PE and KE is crucial for solving pendulum motion problems effectively.
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Homework Statement


A 2.0 m pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the string?


Homework Equations


KE=1/2mv^2
PE=mgh


The Attempt at a Solution


I thought that height would just be 2cos25. Since at the bottom of the pendulum, all PE is converted to KE, PE would equal KE. That makes sense. So my equation was gh=1/2v^2 (Mass cancels). My velocity came out to be .59 m/s and I'm unsure if this is correct or not. Some other sites for this problem talked about height being 2-2cos25. If that is true, why would that be height as opposed to just 2cos25?
 
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gneill said:
attachment.php?attachmentid=41262&stc=1&d=1322276156.jpg

I've Calculated the velocity as 1.92m/s now. The diagram helped, thank you!
 

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