Questions
A body of mass m kg is rotating in a vertical circle at the end of a string of length r metre. The difference in the kinetic energy at the top and the bottom of the circle is
a
mg / r
b
2 mg / r
c
2 mgr
d
mgr
detailed solution
Correct option is C
K.E. at top = 12mv2 K.E. at the bottom = 12mu2Conserving mechanical energy , 12mu2 = 12mv2 +mg2r ⇒12mu2 - 12mv2 = 2mgrTalk to our academic expert!
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