A rod of length L is pivoted at one end & rotated with a uniform angular velocity in a horizontal plane. Let & be the tensions at points L/4 & 3L/4 respectively away from pivoted ends. Then
(a) >
(b) =
(c) <
(d) Relation between & depends on whether the rod rotates clockwise or anticlockwise
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The tension at L/4 =43Mω2[2143L]. The tension at 3L/4 =41Mω2[2141L]. Therefore T1>T2.
I 'll go with (a) :D
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Hey, please show me the thinking or the concept.
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Tension must acount for the centripital acceleration for both the cases. That's all I used with a bit of calculation to compute the centripital acceleration and mass. Am I correct?
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a is correct...:)
C) t1 <t2
http://www.askiitians.com/forums/Mechanics/10/5212/circular-motion.htm
Consider the portion of the rod between L/4 and 3L/4. T1 acts on that portion in a radially inward direction. T2 acts on that portion in a radially outward direction. The net force on that portion must be radially inward. Therefore, T1 > T2.
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You may be correct. But how far I know tension in a segment of rod always acts towards the mid-point of that section. Anyways can you give me a mathematical solution to this problem ? Including the free-body diagram ?(if possible)