A 0.480 kg pendulum bobpasses through the lowest part of its path at a speedof 3
ID: 1743313 • Letter: A
Question
A 0.480 kg pendulum bobpasses through the lowest part of its path at a speedof 3.70 m/s. (a) What is the tension in the pendulum cable at this point if thependulum is 80.0 cm long?N
(b) When the pendulum reaches its highest point, what angle doesthe cable make with the vertical?
°
(c) What is the tension in the pendulum cable when the pendulumreaches its highest point?
N (a) What is the tension in the pendulum cable at this point if thependulum is 80.0 cm long?
N
(b) When the pendulum reaches its highest point, what angle doesthe cable make with the vertical?
°
(c) What is the tension in the pendulum cable when the pendulumreaches its highest point?
N
Explanation / Answer
m = 0.480 kg, v = 3.70 m/s. L = 0.800 m (a) the tension in the pendulum cable at this point = T T - mg = mv2/L T = m(g + v2/L) = 12.9 N b) mv2/2 = mgL(1 - cos) = cos-1[1 - v2/(2gL)] =40.2o c) T = mgcos = 3.60 N
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