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Mathematical Pendulum
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In the case of a pendulum composed of a point mass the potential energy is given by the effect of raising the mass against the gravitational field as the pendulum deviates by a given angle.
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Potential energy of a mathematical pendulum for small angles
Equation
The gravitational potential energy of a pendulum is
| $ U = m g L (1-\cos \theta )$ |
which for small angles can be approximated as:
 $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
The gravitational potential energy of a pendulum with mass
| $ U = m g L (1-\cos \theta )$ |
where
For small angles, the cosine function can be approximated using a Taylor series expansion up to the second term
$\cos\theta\sim 1-\displaystyle\frac{1}{2}\theta^2$
This approximation leads to the simplification of the potential energy to
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
It's important to note that the angle must be expressed in radians.
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Kinetic energy of a mathematical pendulum
Equation
The kinetic energy of a rotating body is given by
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
where $I$ is the moment of inertia and $\omega$ is the angular velocity. For a point mass $m$ rotating at a distance $L$ from an axis, the moment of inertia is
| $ I = m L ^2$ |
hence,
| $ K =\displaystyle\frac{1}{2} m_i L ^2 \omega ^2$ |
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Angular frequency of a mathematical pendulum
Equation
In the case of the mathematical pendulum
the energy can be expressed as
$E=\displaystyle\frac{1}{2}ml^2\omega^2+\displaystyle\frac{1}{2}mgl\theta^2$
and from this expression, we can obtain the angular frequency
| $ \omega_0 ^2=\displaystyle\frac{ g }{ L }$ |
The kinetic energy of the mathematical pendulum with mass $m$, string length $r$, and angular velocity $\omega$ is
| $ K =\displaystyle\frac{1}{2} m_i L ^2 \omega ^2$ |
and the gravitational potential energy is
| $ V =\displaystyle\frac{1}{2} m_g g L \theta ^2$ |
With $\theta$ representing the angle and $g$ the angular acceleration, the equation for the total energy is expressed as
$E=\frac{1}{2}m r^2 \omega^2 + \frac{1}{2}m g r \theta^2$
Given that the period is equal to
$T=2\pi\sqrt{\frac{m r^2}{m g r}}=2\pi\sqrt{\frac{r}{g}}$
we can relate the angular frequency as
| $ \omega_0 ^2=\displaystyle\frac{ g }{ L }$ |
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