Why Is Tension T=2mg and Not T=mg at Time t=τ?

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The discussion centers on the tension in a massless rod attached to a mass m on a frictionless cart shaped like an equilateral triangle. At time t=τ, when the rod is parallel to the ground, the tension is determined to be T=2mg instead of T=mg due to the dynamics of the system as the mass falls. The radial force, expressed as mv^2/L, must equal the tension force, leading to the conclusion that the tension doubles as the mass accelerates. The torque generated by the gravitational force also plays a role in this analysis. Understanding these relationships is crucial for accurately calculating the tension in this system.
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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=τ is T=mv^2/L=2mg and not T=mv^2/L=mg?
 

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What do you think?
Any relevant equations?
Any ideas of you?
 
mv^2/L is the radial force that need to be equated with the tension force...but the torque mgsin 90 = mg
 
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Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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