Question

A stand shows a pendulum in 5 positions. A has the pendulum bob at a 45 degree angle to the left. C has the pendulum bob straight down, and B is halfway between A and C. E is has the pendulum bob at a 45 degree angle to the right, and D is halfway between C and E.
Determine how potential and kinetic energy changes at each position of the pendulum as the ball swings from A to E.

Position A:

Position B:

Position C:

Position D:

Position E:

Answer 1

Assume no friction/air resistance so mechanical energy is conserved. Let the zero of gravitational potential be at C (bob straight down). For a pendulum of length L and angle θ from the vertical, the height above C is h = L(1 − cos θ) and PE = mg h. Here θA = 45°, θB = 22.5°, θC = 0°, θD = 22.5° (right), θE = 45° (right).

Position A (45° left)
- Potential energy: maximum (PE = mgL(1 − cos45°)).
- Kinetic energy: zero (momentarily at rest).
- Total energy = mgL(1 − cos45°).

Position B (halfway down, 22.5° left)
- Potential energy: decreased relative to A (PE = mgL(1 − cos22.5°) ≈ 0.0761 mgL).
This is ≈ 26.0% of the PE at A.
- Kinetic energy: increased, equal to the difference between total energy and PE at B (KE ≈ 74.0% of the A-energy).
- Speed is nonzero and increasing as it moves toward C.

Position C (vertical, 0°)
- Potential energy: minimum (PE = 0 with our reference).
- Kinetic energy: maximum (KE = mgL(1 − cos45°) — all initial PE converted into KE).
- Speed is greatest here.

Position D (22.5° right)
- Potential energy: same as at B (PE ≈ 0.0761 mgL, ≈ 26% of A).
- Kinetic energy: same as at B (≈ 74% of the initial PE), but decreasing as the bob moves toward E.
- Speed is nonzero and decreasing.

Position E (45° right)
- Potential energy: same as at A (PE = mgL(1 − cos45°)).
- Kinetic energy: zero (momentarily at rest).
- Motion reverses direction here.

Summary: PE decreases and KE increases from A to C; at C PE is minimum and KE maximum; from C to E PE increases and KE decreases. Total mechanical energy remains constant (neglecting losses).