Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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:

$$T-mg=ma_c$$

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

share|improve this question
add comment

1 Answer 1

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 $T= mg + {mv^2\over r} = mg + 2mg = 3mg$.

share|improve this answer
    
It is the case that the swing starts off at 90 degrees. –  Kurt Jun 26 '12 at 5:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.