Energy and Work

[Soaring Crane]

A ball is attached to a horizontal cord of length L whose other end is fixed.

a. If the ball is released what will be its speed at the lowest point of its path?

(1/2)mv^2 = mgr, or v^2 = 2gr, v = sqrt(2gr)???

b. A peg is located a distance h directly below the point of attachment of the cord. If h = 0.80 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?

pic:


O_____L________o
+----------------|
-+---------------|
--+--------------|h
---+-------------|----+
----+------------*peg--+
------+----------|----+
---------+----+-O--+

+ = path of circular motion
O = ball

Any help to start is appreciated.

[Leong]

The first one is correct.(r=L).
When the ball reaches its top circular path, it has both kinetic energy and potential energy.
Initial energy,E(initial) = mgL
Total energy when the ball reaches its top circular path,E(final) = \frac{1}{2}mv^2 + mg*2(L-h).
Law of conservation of energy : E(initial) = E(final)