Practice Problem about the Energy of a Pendulum

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The discussion centers on a practice problem regarding the energy of a pendulum, with initial answers provided as 7.35 m/s for speed and 216.09 m/s² for acceleration. Participants emphasize the importance of showing work and reasoning behind the calculations to facilitate better assistance. There is a suggestion to consider energy principles for solving the problem. The conversation highlights the need for detailed explanations rather than just final answers. Overall, the focus is on understanding the underlying concepts of pendulum energy.
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Homework Statement
A small sphere of mass m is fastened to a weightless string of length 0.5m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 60 degrees with the vertical.
a) What is the velocity of the sphere when it passes through the vertical position?
b) What is the instantaneous acceleration when the pendulum is at its maximum deflection?
Relevant Equations
a) v=sqrt(L*g*sine(thet)*tan(thet)
b) ac= v^2/r
Answer:
a) 7.35 m/s
b) 216.09 m/s^2

*Is this correct?
 
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It would be easier to offer you help if you offer a bit more detail for your reasoning.

In both of the cases a) and b), what is your argumentation behind the expressions?
 
stephy said:
*Is this correct?
I suggest you read the forum rules, one of which is that you are required to show work (not just answers).
 
No working out offered by me (yet), but I would look at energy considerations for the first part.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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