# Angular velocity of a simple pendulum

• Felipe Lincoln
In summary: The angular velocity should be equal to the derivative of the position with respect to time, which is given by:##\dot\theta = \omega\,\ddot{x}##

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## Homework Statement

A pendulum with a light rod of length ##l## with a bob of mass ##m## is released from rest at an angle ##\theta_0## to the downward vertical. Find its angular velocity as a function of θ, and the period of small oscillations about the position of stable equilibrium. Write down the solution for θ as a function of time, assuming that ##\theta_0## is small.

## Homework Equations

i) ## x=\theta l##
ii) ## F=-mg\sin\theta##
iii) ## V = mgl(1 - \cos\theta)##
iv) ## K+V=E##

## The Attempt at a Solution

## F## was obtained considering the equation i), the potential was obtained by doing ## F=-\dfrac{dU}{dx}## and them using the equation iv) we get the first answer, which is to find the angular velocity. The result is ##\omega=\pm\sqrt{\dfrac{2g}{l}(\cos\theta - \cos\theta_0)}##.
But what I can't understand is why isn't the angular velocity equal this: ## \omega=\sqrt{g/l}##

Last edited:
A simple sanity check of ##w = \sqrt {\frac g l}##: How can the anglular velocity of a pendulum be a constant? It stops at the peak and is at it's maximum velocity at the bottom. You should also apply a similar sanity check to your calculated result.

Felipe Lincoln said:

## Homework Statement

A pendulum with a light rod of length ##l## with a bob of mass ##m## is released from rest at an angle ##\theta_0## to the downward vertical. Find its angular velocity as a function of θ, and the period of small oscillations about the position of stable equilibrium. Write down the solution for θ as a function of time, assuming that ##\theta_0## is small.

## Homework Equations

i) ## x=\theta l##
ii) ## F=-mg\sin\theta##
iii) ## V = mgl(1 - \cos\theta)##
iv) ## K+V=E##

## The Attempt at a Solution

## F## was obtained considering the equation i), the potential was obtained by doing ## F=-\dfrac{dU}{dx}## and them using the equation iv) we get the first answer, which is to find the angular velocity. The result is ##\omega=\pm\sqrt{\dfrac{2g}{l}(\cos\theta - \cos\theta_0)}##.
But what I can't understand is why isn't the angular velocity equal this: ## \omega=\sqrt{g/l}##
I think you are using the variable ##\omega## for two different things. The equation of motion for simple harmonic motion has the form:
##\theta = A\sin(\omega t +\phi_0)##
Notice that in this equation ##\omega## is a constant rate of change of phase angle ##\phi## with time. This is not the same as ##\dot\theta##.

FactChecker and Felipe Lincoln

## What is the formula for calculating the angular velocity of a simple pendulum?

The formula for calculating the angular velocity of a simple pendulum is ω = √(g/L), where g is the acceleration due to gravity and L is the length of the pendulum.

## How does the length of a simple pendulum affect its angular velocity?

The length of a simple pendulum affects its angular velocity by directly proportional relationship. That means, as the length increases, the angular velocity also increases, and vice versa.

## What factors can affect the angular velocity of a simple pendulum?

The factors that can affect the angular velocity of a simple pendulum include the length of the pendulum, the mass of the bob, the amplitude of the swing, and the force of gravity.

## Is the angular velocity of a simple pendulum constant?

No, the angular velocity of a simple pendulum is not constant. It varies as the pendulum swings back and forth due to changes in the gravitational force and the pendulum's position.

## How is the angular velocity of a simple pendulum related to its period?

The angular velocity of a simple pendulum is inversely proportional to its period. This means that as the angular velocity increases, the period decreases, and vice versa.