# A pendulum 2.20 m long is released (from rest) at an angle theta = 30.0 degrees. Determine the...

## Question:

A pendulum 2.20 m long is released (from rest) at an angle {eq}\theta{/eq} = 30.0 degrees. Determine the tension in the cord at {eq}\theta{/eq} = 15 degrees.

## Conservation of Energy:

The principle of conservation of energy states that for an closed system, the energy before and after an interaction is the same. This means the kinetic energy will be converted solely to potential energy and vice versa. This phenomenon is independent of the path of the process.

Using conservation of energy to solve for the speed, we write:

{eq}\frac{1}{2}mv^2 = mgl{1-\cos{15}} {/eq}, where:

• m is the mass
• v is the speed
• g is gravitational acceleration on Earth
• l is the length of the cord

{eq}\frac{1}{2}v^2 = (9.8)(2.2)(1-\cos15) \\ v = 1.21{/eq}

The radial component of the tension is given as:

{eq}T = m(\frac{v^2}{r} + g\cos{15}){/eq}

Inserting the values of the available parameters, we write:

{eq}T = m(\frac{(1.21)^2}{2.2} + (9.8)\cos{15}) \\ \boxed{T = 10.1m \ N}{/eq}