# Centrefugal Force

centrifugal force + gravity must be along the string to cancel its tension,

Now imagine a right triangle formed by the string (hypotenuse), the vertical leg and the horizontal leg (from the mass to the center of rotation).

The triangle is similar to the one formed by the tension vector (hypotenuse), gravity vector (vertical leg), and centrifugal force vector (horizontal leg).

In this triangle, we have a geometrical relation

(horizontal leg) = (vertical leg) *tan(theta)

or

mv^2/(L sin(theta)) = (m g) tan(theta)

sin (theta) tan(theta) = v^2 /(gL)

Now use the identity

sin (theta) tan(theta) = [1-cos(theta)^2]/cos(theta)

to convert the equation to a quadratic equation in terms of cos(theta), and solve it

cos(theta) = sqrt(1+A^2) – A

where, A = v^2 /(2gL) = 16/2/10/2 = 0.4

theta = 47 degrees

The forces on the mass are its weight “mg” straight down and the string tension “T” up along the string.

Find horizontal (along radius of circle) and vertcal force components;

The horizontal force is TSin(theta) and it provides the centripital accel;

TSin(theta) = mv^2/r

The vertical forces balance, as there is no movement vertically;

TCos(theta) = mg

Divide eqs to ellimiate “T”

Tan(theta) = v^2/gr

This would be easy if you were given the radius “r” of the circle, but you only know the length of the string “L” so use

r = LSin(theta)

Sub. to get;

[Sin(theta)]^2/Cos(theta) = v^2/gL

1 – [Cos(theta)]^2 = (v^2/gL)Cos(theta)

Solve this quadratic eq for Cos(theta);

C^2 + (v^2/gL)C – 1 = 0

Ill let you finish it.