Calculating Work and Potential Energy in a Vertical Circle

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SUMMARY

The discussion focuses on calculating work and potential energy for a ball attached to a thin rod rotating in a vertical circle. The rod has a length of 2.00m, and the ball has a mass of 5.00 kg. The gravitational work done on the ball is calculated using the equation W (gravity) = mgh, where the height h is derived from trigonometric relationships. Participants confirmed that the correct height is approximately 0.27m, leading to a calculated work of around 13 J, which differs from the book's stated answer of 20 J. The discrepancy prompted suggestions to verify calculations and consult instructors for clarification.

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  • Understanding of gravitational potential energy and work-energy principles
  • Familiarity with trigonometric functions and their application in physics
  • Knowledge of basic mechanics, specifically rotational motion
  • Ability to manipulate equations involving height and distance in a circular motion context
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  • Learn how to derive height in circular motion problems using trigonometry
  • Study the work-energy theorem and its applications in physics
  • Explore the concept of conservation of energy in mechanical systems
  • Review common pitfalls in physics problem-solving and how to avoid them
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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of work and energy in rotational dynamics.

Almoore01
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A thin rod of length = 2.00m and negligible mass can pivot about one end to rotate in a vertical circle. A ball of mass m = 5.00 kg is attached to the other end. The rod is pulled aside to angle \theta= 30 degrees and released with initial velocity v = 0 m/s. As the ball descends to its lowest point,

(a) How much work does the gravitational force do on it?
(b) What is the change in the gravitational potential energy of the ball-Earth system?
(c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released?
(d) Doe the magnitude of the answers to (a) through (c) increase, decrease, or remain the same if the angle is increased?



Relevant Equations:
W (gravity) = mgh
W (gravity) = -\DeltaU



So, theoretically, I know what I need to do, but when it comes to plugging in the numbers, I get the incorrect answer. I know that I'm given m, and g is the gravitational constant. What I'm having trouble doing is manipulating the arc of the circle to solve for h, and the h that I keep finding (roughly 0.27 m) gives me the incorrect answer for (a), which should be 20 J. If I'm struggling with part (a), I assume I will have similar issues with the rest of it, so any help would be greatly appreciated.
 
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You problem is that the arc length of the path of the mass is not the same as the change in height.

I found this picture on Google image to help you find the actual height difference:

http://www.physics247.com/physics-homework-help/conservationofenergy6.jpg

The change in height is the distance labeled h in this diagram. Can you use trig and some algebra to find h?
 
I tried that but it wasn't working. What I have is that: 2Rsin((1/2)theta) = l, which is the length from one end to the other end in the diagram you posted. I got 1.035m for that. From there I solved for the length of the perpendicular line in the diagram and got 1.00m. Then I solved for one of the angles in my newly formed triangle and determined that h was 0.2678 m, which when plugged into W (gravity) equation, yields the incorrect answer.
 
OK. The height you found is correct. And using it I get an answer of around 13J for the work done by gravity. Is this close to what your getting? If it is, I would check your numbers.

You have the correct height, so one of the other numbers may be wrong.
 
Yeah, I also get approximately 13 J. The issue is that the back of the book states that the answer is 20 J. Maybe it's wrong, but I've been tearing my head out over this problem...
 
I'm sure that you have the correct number for the height.

I just checked my work and I get the same thing, .27m.

Maybe talk to your instructor about it. These things happen, and I agree that they can be very frustrating.
 
Actually, with a little tweaking, I figured it out. The answer in the book was only slightly off. Thanks a lot for the help.
 

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