The height of a tower is measured by attaching a simple pendulum to its ceiling,
ID: 1653092 • Letter: T
Question
The height of a tower is measured by attaching a simple pendulum to its ceiling, whose length is barely enough to stay off the floor. The pendulum is let go from a small angle, and takes 15 s to return to the same location it started from. a.) How tall is the tower? m If the pendulum mass is let go 0.1 m above the floor, b.) How fast is the mass travelling as it grazes the floor? m/s c.) From what angle (measured from vertical) was this pendulum released? Degrees radians d.) How much % error is incurred in the pendulum's initial angular acceleration by using the small angle approximation? (Remember to use radians here.) % (Does the error in the angular acceleration get worse or better as the pendulum swings towards equilibrium?)Explanation / Answer
T = period = 15 s
h = initial height above the ground = 0.1 m
g = acceleration by gravity = 9.8 m/s²
m = mass of the bob (unknown)
The period of a pendulum is given by:
T = 2 ( L / g )
where
L = length of the string
so
L = gT² / 4²
L = (9.8*(15)2)/(4*3.14*3.14) = 55.9 m < - - - - - - - - - - - - answer a
The potential energy at the initial position equals the kinetic energy at the lowest point.
m v² / 2 = m g h
v = ( 2 g h ) = ( 2 * 9.8 * 0.1 )
v = 1.4 m/s < - - - - - - - - - - - - answer b
The initial angle of the string can be found by:
cos() = (L - h) / L
where
= initial angle
so
= arccos[(L - h) / L] = arccos[(55.9 - 0.1) / 55.9]
= 3.42° in degree < - - - - - - - - - - - - answer c
= 3.42° in radian = 3.42* pi/180 = 0.0596 radian< - - - - - - - - - - - - answer c
% error in initial angular acceleration is
= (0.0596/15)*100 = 0.39 %< - - - - - - - - - - - - answer d
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