Grade 12th passMechanics
Derive velocity of the particle at top bottom and in midway in circular motion.
1 Answer
Vikas TU
At top the minimum speed required to attain the circular motion gives a downward centripital acceleration that is a = g m/s^2.
Hence,
mg = mv^2/R
v = root(gR)
And at the bottom,
0.5mv^2 + mg(2R) = 0.5mVbottom^2
0.5(Vbottom)^2 = 5gR/2
Vbottom = root(5gR).
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