Solve Simple Loop Problem: Find Speed at Point A

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The discussion centers on calculating the speed of a bead at the top of a loop-the-loop, released from a height of 3.75R. The initial approach using the formula a = v^2/r was deemed incorrect, prompting a suggestion to apply conservation of energy principles instead. The bead's potential energy at the height converts to kinetic energy at point A. The correct method involves equating the potential energy at the starting height to the kinetic energy at point A to find the speed in terms of R and g. Utilizing conservation of energy provides the accurate solution for the bead's speed at point A.
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Homework Statement


A bead slides without friction around a loop-the-loop as shown in the figure below. The bead is released from a height h = 3.75R.

(a) What is the bead's speed (V) at point A? Answer in terms of R and g, the acceleration of gravity.

I don't have the picture it's in the book so I'll describe it. It's a downward slope that go's into a loop. 'A' is at the the top of the loop and R is the loop's radius.

Homework Equations



a=v^2/r

The Attempt at a Solution



Since it's perfectly parallel with the y axis, and therefor i assume a is 9.81, I get (gr)^1/2, but my webassignment program says that's wrong.
 
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Guruu said:
A bead slides without friction around a loop-the-loop as shown in the figure below. The bead is released from a height h = 3.75R.
…
(a) What is the bead's speed (V) at point A? Answer in terms of R and g, the acceleration of gravity.
…
a=v^2/r
…
Since it's perfectly parallel with the y axis, and therefor i assume a is 9.81, I get (gr)^1/2, but my webassignment program says that's wrong.

Hi Guruu! :smile:

What does a=v^2/r have to do with part (a)? :confused:

Hint: use conservation of energy. :smile:
 
yup conservation of energy is the way to go
 

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