physics
posted by Kalen .
We are learning about circular motion, uniform and non uniform. how do you find tension in a string, when the string is attached to a object with mass m and length r? What about if you are swinging it around in a horizontal circle and the string makes a angle theata with the horizontal?

physics 
MathMate
If we ignore gravity, and calculate the tension in the string due solely to the circular motion, then
Tc=mrω²
where
Tc=tension due to circular motion
m=mass attached to the end of the string
r=radius
&omega=angular velocity, in radians / second
For a horizontal circular motion, the above value of Tc should be vectorially added to the weight, i.e.
Total Tension, T
= &radic(T²+(mg)^2)
The angle
tan^{1}(mg/Tc)
is the angle with the horizontal.
For a vertical circular motion, Tc must be added to the vertical component of weight, mg (positive upwards)
The vertical component can be obtained by mg*sin(θ),
where θ=angle above horizontal, equals zero when mass is horizontal and going up.
Tension T
=Tc  mg*sin(θ) 
physics 
bobpursley
The only thing I would add to the above, is that the r in the formula for centripetal force is the radius of the horizonal circle, not the length of the string. if l is the length of the string, then r=lcosTheta. This makes Tc = mw^2 l cosTheta

physics 
MathMate
Thank you Bob.

physics 
Kalen
In the total tension, what does &radic mean?
Total Tension, T
= &radic(T²+(mg)^2)
The angle
tan1(mg/Tc)
is the angle with the horizotal 
physics 
MathMate
Total Tension, T
= √(T²+(mg)^2)
The angle
tan1(mg/Tc)
is the angle with the horizotal
√ is the equaivalent of squareroot.
The formula is a typical application of Pythagoras theorem while adding the vertical and horizontal components of force. I could rewrite it as:
Total Tension, T
= sqrt(T²+(mg)^2)
The angle
tan^{1}(mg/Tc)
is the angle with the horizotal
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