Question

A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.4 m/s and the tension in the rope is T = 15.1 N. Now a peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. Return to the original mass. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

Answer 1

Let the rope length be L and the mass be m.

1) From energy (release horizontal to bottom): 1/2 m v^2 = m g L, so
L = v^2/(2g).

2) At the bottom the tension is T = mg + m v^2/L. Using v^2/L = 2g from (1) gives
T = 3 m g => m = T/(3g).

3) The peg at 4/5 of the way down means cosθ = (4/5) for the position "directly to the right of the peg." The speed there from energy is
v_θ^2 = 2 g L cosθ.

4) Radial force balance: T_θ = m v_θ^2/L + m g cosθ = m(2g cosθ) + m g cosθ = 3 m g cosθ.

Substitute m = T/(3g) and cosθ = 4/5:
T_θ = 3 (T/(3g)) g (4/5) = T (4/5).

Numerically: T_θ = 15.1*(4/5) = 12.08 N ≈ 12.1 N.

Answer: 12.1 N.