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Answer to Question #20277 in Mechanics | Relativity for zk
Question #20277
Q: A body weighing 2 kg tied at the end of string 1 meter long and rotating in a vertical circular path at a speed of 6 meter per second. What is the tension at the string at top and at bottom?
Expert's answer
T = mg + ma
T - the tension at the string
a - the acceleration of the body
a = v^2/r
On the top:
T = ma - mg = m(v^2/r - g) = 2*(6^2/1-10) = 52 H
On the bottom:
T = ma + mg = m(a + g) = 2*(36+10) = 92 H
T - the tension at the string
a - the acceleration of the body
a = v^2/r
On the top:
T = ma - mg = m(v^2/r - g) = 2*(6^2/1-10) = 52 H
On the bottom:
T = ma + mg = m(a + g) = 2*(36+10) = 92 H
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