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Discussion Response Post to:
Simple Triky question
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Mechanics
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Author
Message
1 Mar 2007 18:06:59 IST
Subject:
Re:Simple Triky question
truly
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assume it to be hanged at point A which is at distance "x" from the center. MI about this point A will be = (mr^2)/2 + mx^2 (say this is denote by MI)
whn displaced a small angle
, the restoring torque = mgx sin
thus MI d
2
/ d
2
t = mgx sin
= mgx
(as sin
=
)
d
2
/ d
2
t = (mgx / (mr^2)/2 + mx^2 )
d
2
/ d
2
t = P
whr P = (mgx / (mr^2)/2 + mx^2 )
Time period T = 2 pi root (1/P)
for T min P shud be max now u can use maxima minima
Truly Mittal
B.Tech IIT Guwahati
trulymittal@gmail.com
this reply: 5
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