- #1
Maximum Angular Velocity of a quarter circle
- Thread starter Yasin
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Discussion Overview
The discussion revolves around finding the maximum angular velocity of a quarter circle, focusing on the application of energy conservation principles and the dynamics of a physical pendulum. Participants explore various approaches to derive the formula and clarify the underlying physics.
Discussion Character
- Homework-related
- Technical explanation
- Debate/contested
Main Points Raised
- One participant states the answer for maximum angular velocity is 0.839*(g/b) but expresses confusion over its derivation.
- Another participant emphasizes the need for effort in problem-solving before seeking help.
- A participant attempts to relate gravitational potential energy to kinetic energy, initially deriving a formula but later realizes it may not apply to the quarter circle scenario.
- Questions arise about the distance the center of mass falls, with some participants suggesting it cannot be 'b' for the entire semicircle.
- There is a correction regarding the potential energy calculation, indicating that the center of mass does not fall a distance 'b'.
- One participant claims to have solved the problem and suggests the textbook solution is incorrect, sharing their findings in a picture.
- A later reply critiques the initial solution, suggesting that the problem involves a physical pendulum and that the effective radius for calculations is not 'b'.
Areas of Agreement / Disagreement
Participants express differing views on the correct approach to the problem, particularly regarding the application of energy conservation and the distance the center of mass falls. No consensus is reached on the correct solution or the validity of the textbook answer.
Contextual Notes
There are unresolved assumptions about the system's dynamics, including the effective radius and the interpretation of potential energy in the context of a quarter circle. The discussion reflects varying interpretations of the physics involved.
- #2
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You have to show some effort before you ask for help. The answer itself is not a relevant equation.
- #3
Yasin
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Sorry, I am new and my attempt I tried converting mgh(gravitational potential energy) into (mv^2)/2 (Kinetic energy)
b being the height i get mgb=mv^2/2 so i get 2gb=v^2
v^2/r^2 is w^2 so i get 2gb/b^2=w^2 which is sqr(2g/b) as the answer
but somehow the answer in the book i use to study is different so i must be wrong
Now i see this is wrong because this is valid for a string and the question has a segment quarter circle.
Same method but using point mass at line where angle is 45 degrees i now get
w=sqr(2g(1-sin45)/b)
b being the height i get mgb=mv^2/2 so i get 2gb=v^2
v^2/r^2 is w^2 so i get 2gb/b^2=w^2 which is sqr(2g/b) as the answer
but somehow the answer in the book i use to study is different so i must be wrong
Now i see this is wrong because this is valid for a string and the question has a segment quarter circle.
Same method but using point mass at line where angle is 45 degrees i now get
w=sqr(2g(1-sin45)/b)
Last edited:
- #4
CWatters
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Regarding the potential energy... Does the whole semicircle fall a distance b?
- #5
Yasin
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The fall is any distance but here I assume when the part where 'm' is located is nearest to bottom that will be the max speed thus max angular velocityCWatters said:
- #6
CWatters
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What I was hinting at is the available PE cannot be mgb because the centre of mass cannot fall a distance "b".
- #7
Yasin
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Yes you are right and i did change the last one by editing the first postCWatters said:
and i think i have solved it thanks for all support the solution is as follows in picture for any interested
Some reason the textbook solution is wrong
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- #8
- 15,946
- 9,141
Mechanical energy conservation is the way to go, but your solution is incorrect. Here you have an extended object that is acted upon by the external force of gravity. If you were to pretend that the entire mass of this contraption were concentrated at one special point, where would that special point be? Once you find that, then you can say ##\frac{1}{2}mv^2=\frac 1 2 m (\omega r)^2## where ##r## is not ##b##, but the distance from the pivot to the aforementioned special point. What you have here is a physical pendulum.
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