A rod of length L is pivoted at one end & rotated with a uniform angular velocity in a horizontal plane. Let **\(T_{1}\)** & **\(T_{2}\)** be the tensions at points L/4 & 3L/4 respectively away from pivoted ends. Then

(a) **\(T_{1}\)** > **\(T_{2}\)**

(b) **\(T_{1}\)** = **\(T_{2}\)**

(c) **\(T_{1}\)** < **\(T_{2}\)**

(d) Relation between **\(T_{1}\)** & **\(T_{2}\)** depends on whether the rod rotates clockwise or anticlockwise

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## Comments

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TopNewestThe tension at

L/4\(=\frac{3}{4}M\omega^2[\frac{1}{2}\frac{3}{4}L]\). The tension at3L/4\(=\frac{1}{4}M\omega^2[\frac{1}{2}\frac{1}{4}L]\). Therefore \(T_1 > T_2\).Log in to reply

http://www.askiitians.com/forums/Mechanics/10/5212/circular-motion.htm

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C) t1 <t2

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\(a\) is correct...:)

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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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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)

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