A 0.420 kg pendulum bobpasses through the lowest part of its path at a speedof 3

Question

A 0.420 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

a.) T - mg = mv^2/r, where v = velocity of the boband r = cable length.
=> T
= m(g + v^2/r)
= (0.42)[9.8 + ((3.7)^2/0.8)]
= 11.3 N

b.)
(2r)*sin^2(/2) = v^2/2g
sin(/2) = v / 2(rg)
sin(/2) = 3.7 / 2(0.8)(9.8)
sin(/2) = 0.6607
= 82.705°

c.) At the highest point, its velocity is zero. So no centripetalforce on it. Hence, T should equal the component of mg in itsdirection.
T = mg cos 82.706° = (0.42)(9.8) (0.127)
T = 0.5227 N

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