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Physic (store energy)

Question Four
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the
gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces
it 10 o
from the equilibrium position, and releases it. Using a stopwatch, the student determines that the
period of the pendulum is 1.44 s. Determine the experimental value of the gravitational acceleration in
two decimal places and the angular speed of the pendulum. (10 marks)

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Jay Peels

  1. angular speed will vary in the arc. You could get average angular speed.

    Period=2PIsqrt(L/g)
    L/g=(period/2PI)^2

    g= Length/(period/2PI)^2

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    bobpursley

  2. T = 2pi * sqrt L/g,
    T^2 = 39.5*L/g = 1.44^2,
    39.5 * 0.51/g = 2.074,
    20.145/g = 2.074,
    g = 9.71 m/s^2.

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    Henry

  4. No fair using the answer :)

    How far does the pendulum drop from the ten degrees?
    h = .510 (1 - cos 10) = 0.00775 meters drop
    so potential energy at top = m g h = m g (0.00775)

    10 degrees * 2 pi radians/360 degrees = .1745 radians
    period = 1.44 s
    say theta = .1745 sin w t = .1745 sin 2 pi t/T = .1745 sin (2 pi t/1.44)
    = .1745 sin (4.36 t)
    angular speed = d theta/dt = omega = 0.761 cos (4.36 t)

    so max speed at bottom = 0.761 * .510 = .388 m/s
    (1/2) m v^2 = m g h which we know is 0.00775 m g
    so
    g = (.5*.388^2)/0.00775 = 9.72 m/s^2

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    Damon - hey no fair :)

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