Suppose the simple pendulum was released from an angle of θ = 53° , with L = 0.75 m and m = 0.13 k?
Suppose the simple pendulum shown below were released from an angle of θ = 53° from horizontal, with L = 0.75 m (the length of the string holding the bob) and m = 0.13 kg
A) What would be the speed of the bob at the bottom of the swing?
B)To what height would the bob swing to on the other side?
C)What angle of release would give half the speed of that for the 53° release angle at the bottom of the swing?
There are no pendulum examples in my book, so I'm completely lost on this one
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This is about energy conservation.
Calculate the height above the low point for an angle of 53º. That is h = 0.75 – 0.75 cos 53. Calculate the potential energy for that height. At the bottom of the swing that is all KE, so you can use that to calculate the speed. You do not need the mass as that cancels out. (a)
½mV² = mgh
V² = 2gh
b) same as part a
c) you have the speed in part a, divide by 2, then work backwards to the angle.
Kinetic Energy in J if m is in kg and V is in m/s
KE = ½mV²
Potential Energy in J
PE = mgh
g is the acceleration of gravity 9.8 m/s²
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