Angular velocity of a simple pendulum

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}##
  • #1

Felipe Lincoln

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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}##
 
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  • #2
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.
 
  • #3
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##.
 
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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.

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