A mass hangs on the end of a massless rope. The pendulum is held horizontal and

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 = 2.7 m/s and the tension in the rope is T = 21.3 N.

1. 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.

How fast is the mass moving when it is at the same vertical height as the peg (directly to the right of the peg)?

2. 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)?

Explanation / Answer

1.

at the bottom . force equation is given as

T - mg = mv2/L

21.3 - mg = m (2.7)2/L

using conservation of energy

initial potential energy = final potential energy + kinetic energy

mgL = mg(L/5) + (0.5) m v2

gL = g(L/5) + (0.5) v2

g (8L/5) = v2

v = sqrt(g (8L/5))

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