Angular velocity of a simple pendulum

  • #1
Felipe Lincoln
Gold Member
99
11

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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Answers and Replies

  • #2
FactChecker
Science Advisor
Gold Member
6,051
2,336
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
tnich
Homework Helper
1,048
336

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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