Revolving Ball on String Problem

[PhysicsinCalifornia]

Another physics question for all you smart people out there

A 0.50-kg ball that is tied to the end of a 1.5-m light cord is revolved in a horizontal plane with the cord making a certain degree with the vertical. If the ball is moving at 4.0 m/s, how can I find the angle that the cord makes with the vertical?

This one is a toughie!

I worked on this a couple of hours, but I keep on running into the same dead ends, could anyone give us some pointers?

(This is the direction I've been trying to go with no luck)

Fx = Tsin(theta) = m((v^2)/r)
Fy = Tcos(theta) - Fg = 0

 

[Berislav]

What does r stand for in your equation?

 

[PhysicsinCalifornia]

r stands for the radius

 

[Berislav]

r stands for the radius

Well, of course. I meant what value does r take in your equation.

EDIT: Because, you know, it's not 1.5 m.

 

[Doc Al]

(This is the direction I've been trying to go with no luck)

Fx = Tsin(theta) = m((v^2)/r)
Fy = Tcos(theta) - Fg = 0

So far, so good. As I think Berislav was trying to point out, r depends on \theta: rewrite r in terms of the length and the angle. Also, realize that F_g = mg.

Then just eliminate T and solve for the angle. Hint: You may need to use a trig identity.

(Note: We can't tell where you got stuck, since you only showed the first step in your solution.)

 

[PhysicsinCalifornia]

Well,

I got up to where

tan(theta)sin(theta) = 1.088

I just need to find (theta) now.

Any help?

Thanks

 

[Doc Al]

Maybe this will help:
\sin^2\theta / \cos\theta = (1 - \cos^2\theta)/ \cos\theta

 

[PhysicsinCalifornia]

Yea, that was my next step.

**New info**

I was informed by my friend that I should use the half-angle identity for this!

I don't know how to answer this problem yet, but I'm getting there

 

[Doc Al]

Use what I gave you in post #7 and solve for \cos\theta. (Rewrite the expression as a quadratic equation in \cos\theta, then solve it.)