The center of mass of the pendulum bob is a distance h 1 =130.0 cm above the tab

Question

The center of mass of the pendulum bob is a distance h1=130.0 cm above the table when the bob is at the equilibrium position. The pendulum bob is pulled back and released from a point where the bob´s center of mass is at a height of h2=168.0 cm above the table.

A) What will be the speed of the pendulum bob as it reaches the bottom of the swing. Use g = 9.8 m/s2.

B) If the length L of the pendulum is 74.6 cm, what is the angle ?0 from which the bob is released? First write down a symbolic expression in terms of given parameters L, g, h1, and h2

Explanation / Answer

a) From law of conservation of energy

m g h1 + 1/2 m v12 = m g h2 + 1/2 m v22

1/2 m v22 - 1/2 m v12 = m g h1 - m g h2

v22 - v12 = 2 g (h1 - h2)

v2 = sqrt [v12 + 2 g (h1 - h2)]

= sqrt [2 * 9.8 (1.68 - 1.30)]

speed of the pendulum bob = 2.73 m/s

b) v = sqrt [2 g l * (1 - cos theta)]

1 - cos theta = v2 / 2 g L

cos theta = 1 - [v2 / 2 g L]

= 1 - [2.732 / 2 * 9.8 * 0.746]

theta = 60.6 deg

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