Solve Simple Loop Problem: Find Speed at Point A

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SUMMARY

The discussion centers on calculating the speed of a bead at point A of a loop-the-loop, released from a height of h = 3.75R. The correct approach involves applying the conservation of energy principle rather than relying solely on the centripetal acceleration formula a = v^2/r. The bead's speed at point A can be derived using gravitational potential energy and kinetic energy equations, leading to the conclusion that V = √(g * (3.75R - R)) = √(2.75gR).

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  • Understanding of conservation of energy principles in physics
  • Familiarity with centripetal acceleration concepts
  • Basic knowledge of gravitational potential energy and kinetic energy equations
  • Ability to manipulate algebraic expressions involving variables R and g
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  • Study the conservation of mechanical energy in closed systems
  • Learn how to apply centripetal force equations in circular motion problems
  • Explore the relationship between potential energy and kinetic energy in physics
  • Practice solving loop-the-loop problems with varying heights and radii
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Students studying physics, particularly those focusing on mechanics and energy conservation, as well as educators seeking to clarify concepts related to circular motion and energy transformations.

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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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