A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The pendulum is pulled to one side so that the rod is at an angle θ from the vertical, and released from rest.

a. Show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere?
b. Repeat part (a) for the instant when the pendulum rod is at an angle 9/2 from the vertical.
c. Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

amplitude

hope this