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A massless rod of length L attached to mass m and with axle to cart of mass M. The cart has a shape of equilateral triangle (edge L). the cart is at rest and its center of mass is above x=0 and a rod is perpendicular to the ground. the cart is free to move without friction. at time t=0 the mass m is released and starts to fall to the left. at time t=τ the rod is parallel to the ground. Why the tension from the stick at time $t=\tau$ is $T=mv^2/L=2mg$ and not $T=mv^2/L=mg$? enter image description here

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closed as off-topic by John Rennie, Dilaton, Emilio Pisanty, Brandon Enright, jinawee Feb 28 '14 at 16:25

This question appears to be off-topic. The users who voted to close gave this specific reason:

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@ Wojciech What is the tension that the cart feels? – user40987 Feb 28 '14 at 11:47
I think I've seen a different question on this case a couple days ago. Is it homework? Also please elaborate on the answers You provide, why You think it should be one and not another. – Wojciech Feb 28 '14 at 12:03
mv^2/L is the radial force that need to be equated with the tension force....but the torque mgsin 90 = mg..I have no idea why it is 2mg may you give an explanation? – user40987 Feb 28 '14 at 12:16
up vote 0 down vote accepted

Ok, the tension will be equal to centrifugal force $T=mv^2/L$. Try to calculate velocity from energy conservation law $mgL=mv^2/2$ and then see what happens.

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thank you very much – user40987 Feb 28 '14 at 12:32

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