Speed of Ball Attached to Horizontal Cord

[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)

 

[wais]

a. The speed of the ball at the lowest point of its path can be calculated using the equation (1/2)mv^2 = mgr, where m is the mass of the ball, v is its speed, g is the acceleration due to gravity, and r is the distance from the point of attachment to the lowest point of the ball's path. Since the ball is released from rest, its initial velocity is 0, and the equation simplifies to v^2 = 2gr. Plugging in the values for g (9.8 m/s^2) and r (L, the length of the cord), we get v = sqrt(2gr). This means that the speed of the ball at the lowest point of its path will depend on the length of the cord.

b. To calculate the speed of the ball when it reaches the top of its circular path about the peg, we can use the conservation of energy principle. At the top of the path, the ball has both kinetic energy (due to its speed) and potential energy (due to its height above the ground). The total energy is equal to the sum of these two energies. We can set this equal to the energy at the bottom of the path, which is only kinetic energy (since the ball is released from rest). The equation is written as (1/2)mv^2 + mgh = (1/2)mv^2, where m is the mass of the ball, v is its speed, g is the acceleration due to gravity, h is the height of the ball above the ground, and r is the radius of the circular path. We can rearrange this equation to solve for v, which gives us v = sqrt(2gh). Since h = 0.80L, we get v = sqrt(2g(0.80L)) = sqrt(1.6gL). This means that the speed of the ball when it reaches the top of its circular path about the peg will also depend on the length of the cord.