# Weight Attatched to string spun in horizontal circle: Find angle with horizontal

1. Sep 1, 2009

### DmJeffer

1. The problem statement, all variables and given/known data
A 104 g stone is whirled in a horizontal circle on the end of an 83 cm long string. The stone takes 1.2 s to make each complete revolution. Determine the angle that the string makes with the horizontal.

2. Relevant equations
f=ma
v=(2*pi*r)/T
Centripetal acceleration=V^2/r

3. The attempt at a solution
I'm assuming I need to find the centripetal acceleration necessary to maintain circular motion. I then need to find the tension of the string and the normal force, and then find the angle needed to produce the needed centripetal acceleration.

r=.83sin(θ)
v=(2*.83sin(θ)*pi)/1.2
A=((2*.83sin(θ)*pi)/1.2)^2/(.83sin(θ))

And this is where I get stuck. I feel like the problem should be just be simple algebra and substitution, but I am having trouble finding an equation for the normal force. Am I on the right track with this?

Thanks for any assistance.
Dmjeffer

2. Sep 1, 2009

### rock.freak667

ω=2π/T, so you can find ω, and hence the centripetal force using mω2r.

The horizontal component of the tension provides the centripetal force and the vertical component counteracts the weight.

3. Sep 1, 2009

### jambaugh

It will help to draw a vector diagram (r,z) with gravity pulling downward (f_z = mg = m * 9.8 m/s/s) and the calculated centripetal acceleration and force acting horizontally f_r).

Remember that since the axis of rotation is vertical the centripetal force is perpendicular to it and thus horizontal.

You know the resulting vector must be in line with the string (so the tension can exactly cancel). You should then be able to write one or two equations and solve for the angle.

4. Sep 1, 2009

### DmJeffer

Thanks for the fast response.

So, now I have

ω=5.2

centripetal force=(.104)(5.2)^2(.83cos(θ))=2.3cos(θ)

Tcos(θ)=2.3cos(θ) : This is the horizontal component of the tension force equal to the centripetal force

T=2.3

Tsin(θ)=9.8m/s/s*.104kg : This is the vertical component of the tension equal to the force of gravity

Solving for this, I get an angle of 26 degrees, which is apparently wrong. Am I going in the right direction with this? Or am I still misunderstanding the problem?

5. Sep 1, 2009

### rock.freak667

It does look like you did everything correct. It could be that you need to work with more decimal places to get a more accurate answer.

6. Sep 1, 2009

### DmJeffer

Exactly right, I sincerely appreciate the help, this problem has been bugging me for a few days now.