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I am given a series of three equations, which someone has used to determine the Tension force in a rope swing:

$$mgh=1/2mv^2$$ $$T-mg=\frac{mv^2}{r}$$ $$T=3mg=1764N$$ I am asked to exaplain the steps of the calculations, and the concepts involved. To begin, I realize the first equation is stating that the Potential Energy of the person swinging on rope will be equal to their Kinetic energy, at bottom or swing, according to the conservation of energy laws. Additionally, setting up this equation allows us to calculate the $v$ of the person swinging on rope, which is needed in the second equation. The second equation states that the Tension in the rope, at bottom of swing, will be equal to the sum of the Centripetal Force and weight of person swinging on rope. But I am having trouble understanding the last equation. Clearly, a substitution has been made on one side of the equation to arrive at $3mg$.

I know that I can sub Centripetal acceleration into the equation:


Could I have a hint to guide me in the right direction? Are there several substituions involved to arrive at $3mg$.

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up vote 1 down vote accepted

The assumption is that the swing starts off at 90 degrees, so that r=h, otherwise it isn't true. Then ${mv^2\over 2} = mgr$, so ${mv^2\over r} = 2mg$, so $T= mg + {mv^2\over r} = mg + 2mg = 3mg$.

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It is the case that the swing starts off at 90 degrees. – Kurt Jun 26 '12 at 5:29

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