Question

A 0.440-kg pendulum bob passes through the lowest part of its path at a speed of 3.16 m/s.

(a) What is the magnitude of the tension in the pendulum cable at this point if the pendulum is 79.0 cm long?

(b) When the pendulum reaches its highest point, what angle does the cable make with the vertical? (Enter your answer to at least one decimal place.)


(c) What is the magnitude of the tension in the pendulum cable when the pendulum reaches its highest point?

Answer 1

Answer: Given mass of pendulam m=0.440kg

Speed of pendulam v= 3.16 m/s

Length of pendulam L= r= 79.0 cm= 0.79m

If the bob were at rest the tension will be simply equal to the weight w=mg.
Since it is moving with a velocity v, the tension should provide the centripetal force of ( mv^2 / r ) also.
T = m v^2 / r + mg = m {[v^2 /r] + g}
= 0.440{[3.16^2 / 0.79] + 9.8} = 9.87 N

The magnitude of the tension in the pendulum cable at this point T= 9.87N
b] The vertical height through which it goes up is given by
h = v^2 /2g

and the angle is given by X

cos X = [L -h] / L

cos X = {0.79 - [3.16^2/ 19.6]} / 0.79
X = 69.2°

angle does the cable make with the vertical is X= 69.2 degrees
c) the magnitude of the tension in the pendulum cable when the pendulum reaches its highest point is
T = mg cos X = 0.44*9.8*cos 69.2

T = 1.53 N