A thin rod of length 0.79 m and mass 130 g is suspended freely from one end. It is pulled to one side and then allowed to swing
like a pendulum, passing through its lowest position with angular speed 3.61 rad/s. Neglecting friction and air resistance, find (a) the rod's kinetic energy at its lowest position and (b) how far above that position the center of mass rises.
1 answer:
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Answer:
0.15694 m
Explanation:
m = Mass of block =
v = Velocity of block = 11.2 m/s
k = Spring constant = 110 N/m
Here the kinetic energy of the fall and spring are conserved
The amplitude of the resulting simple harmonic motion is 0.15694 m