Understanding Centripetal Force and its Relation to Circular Motion

[Destrio]

A 1.34kg ball is attached to a ridig vertical rod by means of two massless strings each 1.70m long. The strings are attached to the rod at points 1.70m apart. The system is rotating about the axis of the rod, both strings being taut and forming an equilateral triangle with the rod,. The tension in the upper string is 35.0N.
a) Find the tension in the lower string
b) Calculate the net force on the ball.
c) What is the speed of the ball.

I did part a, found the tension of the lower string to be 8.717N
For part B I know that Fnet in the y direction is = 0
and Fnet = Fc
The forces in the x direction are 30.11N in the top string, and 7.549N in the lower string
I'm not sure how to proceed from here

Thanks

 

[drpizza]

I'm going to take for granted that your results thus far are correct.

You've pretty much got it - You've got those two forces in the x-direction (directed toward the center of rotation.) What kind of acceleration is this that we're talking about?

And, once you know how large this force is, I would assume you know
$$F_c=mv^2$$

 

[Destrio]

Would Fc = my two x tension forces?
so:
Fc = 30.11N + 7.549N = 37.659N

Then i could solve for speed
Fc = mv^2
37.659N = 1.34kg * v^2
v = 5.310m/s

 

[Destrio]

am i correct in that thinking?

 

[FedEx]

$$F_c=mv^2$$

How?

I think that we will have to resolve the tensions in the two strings.

 

[drpizza]

Destrio, yes. (Again, I'm assuming your calculations are correct for the x and y components)

 

[FedEx]

Destrio, yes. (Again, I'm assuming your calculations are correct for the x and y components)

Are you trying to say that F_c is the centripetal force. How is centripetal force equal to mv^2. It is mv^2/r, isn't it?