Saturday

February 28, 2015

February 28, 2015

Posted by **Kalen** on Wednesday, October 7, 2009 at 12:19pm.

- physics -
**MathMate**, Wednesday, October 7, 2009 at 12:41pmIf 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**, Wednesday, October 7, 2009 at 12:43pmThe 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**, Wednesday, October 7, 2009 at 12:46pmThank you Bob.

- physics -
**Kalen**, Wednesday, October 7, 2009 at 1:24pmIn the total tension, what does &radic mean?

Total Tension, T

= &radic(T²+(mg)^2)

The angle

tan-1(mg/Tc)

is the angle with the horizotal

- physics -
**MathMate**, Wednesday, October 7, 2009 at 1:29pmTotal Tension, T

= √(T²+(mg)^2)

The angle

tan-1(mg/Tc)

is the angle with the horizotal

√ is the equaivalent of square-root.

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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