- #1
trilerian
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A pendulum of length 0.6m swings so that at the bottom of its swing, it contacts a peg located 0.4m from the pivot point (0.2m from the CG). At what angle from vertical must the bob be released so that it will just make a revolution around the peg?
F = Mg = M*V^2/r
Mg (l-l cos x) = 1/2 MV^2
g (l-l cos x) = 1/2 gr
l (1- cos x) = 1/2 r
1- cos x = r/2l
1-r/2l = cos x
x = cos^-1 (1- .2/(2*.6) = 33.6
I tried an improvised impromptu experiment and wasn't getting anywhere near the angle 33.6 to work.
F = Mg = M*V^2/r
Mg (l-l cos x) = 1/2 MV^2
g (l-l cos x) = 1/2 gr
l (1- cos x) = 1/2 r
1- cos x = r/2l
1-r/2l = cos x
x = cos^-1 (1- .2/(2*.6) = 33.6
I tried an improvised impromptu experiment and wasn't getting anywhere near the angle 33.6 to work.