A simple pendulum is shown - a mass of 2.5 kg suspended from the ceiling by mean

Question

A simple pendulum is shown - a mass of 2.5 kg suspended from the ceiling by means of a string of length 2.9 m. Assume that the acceleration due to gravity is g = 10 m/s2, and that there is no friction or air resistance.

(a) Suppose you were to release the pendulum from rest, starting from an angle = 31.8 degrees with respect to the vertical, as shown. What will be the speed of the pendulum at the instant it swings through its lowest point (that is, when it's momentarily hanging vertically)? Answer in m/s

(b) Assuming the same situation as in part (a), what will be the speed of the pendulum at the instant it reaches an angle of 30 degrees above the vertical? Answer in m/s

c) Assume you start with the pendulum again at 31.8 degrees with respect to the vertical, but rather than releasing it from rest, you give it a push downward. It swings to the other side, and reaches a maximum angle of 78.2 degrees with respect to the vertical. What must have been the initial speed of the pendulum just after you pushed it? Answer in m/s

d) Suppose you start with the pendulum hanging vertically, at rest. You then give it a push so that it starts swinging with a speed of 1.2 m/s. What maximum angle (in degrees) will it reach, with respect to the vertical, before falling back down? Answer in degrees

e) Same situation as in part (d), but now, due to friction at the pivot point, 1 Joule of the pendulum's initial kinetic energy is lost during the upward swing. Then again, What maximum angle (in degrees) will it reach, with respect to the vertical, before falling back down? Answer in degrees

Explanation / Answer

a) PE lost = mgh = 2.5*9.8*2.9*(1 - cos(31.8)) = 10.66 J
By using conservation of energy
10.66 = 0.5mv^2
v = 2.92 m/s
b) By using conservation of energy
10.66 = 2.5*9.8*2.9(1 - cos(30)) + 0.5*2.5*v^2
v = 0.955 m/s
c) By using conseration of energy
2.5*9.8*2.9(1 - cos(78.2)) = 2.5*9.8*2.9(1 - cos(31.8)) + 0.5*2.5*v^2
v = 6.057 m/s
d) 0.5*2.5*1.2^2 = 2.5*9.8*2.9*(1 - cos(theta))
theta = 12.92 deg
e) 0.5*2.5*1.2^2 - 1= 2.5*9.8*2.9*(1 - cos(theta))
theta = 8.60 deg

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