Solving the Tension Equation: Understanding the Progression

[msudawgs267]

I have a question about this equation (link below). I am working on a tension problem and I don't understand how you go from Sum of F= Ft-mg to
v= [square root of] (Ft-mg)r /m

The progression from one to the other is confusing me. I don't see how can use (mg) in one part of the equation and in the next part it is (v)squared and the back to (mg) at the end.


http://img3.imageshack.us/img3/4830/equationi.jpg


I hope my question was not written too unclear and any help would be amazing. Thanks!

 

[alphysicist]

Hi msudawgs267,

I have a question about this equation (link below). I am working on a tension problem and I don't understand how you go from Sum of F= Ft-mg to
v= [square root of] (Ft-mg)r /m

The progression from one to the other is confusing me. I don't see how can use (mg) in one part of the equation and in the next part it is (v)squared and the back to (mg) at the end.

I would not say the expression on the left is progression from one step to another. The starting point is Newton's law in the form:

$$\sum F = m a$$

and then they are saying that

$$\sum F \to F_T - mg$$

and

$$m a \to m\frac{v^2}{r} \ \ \ \mbox{(for centripetal acceleration)}$$

Setting these two new expressions equal to each other and solving for v gives the answer.