a pendulum, 2.0 m in length, is released with a push when the string is at an angle of 25 o from the vertical. if the initial sp eed of the pendulum is 1.2 m/s, what is its speed at the bottom of the swing?
1 answer:
Answer:
1 / 2 m v^2 = L m g (1 - cos θ)
This is the KE due to the pendulum falling from a 25 deg displacement
v^2 = 2 L g (1 - cos 25) = 2 * 2 * 9.8 (1 - .906) = 3.67 m^2/s^2
v = 1.92 m/s this is the speed due to an initial displacement of 25 deg
Its speed at the bottom would then be
1.92 + 1.2 = 3.12 m/s since it gains 1.92 m/s from its initial displacement
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