Angular acceleration of Pendulum equation

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SUMMARY

The equation for the angular acceleration of a pendulum is definitively given by θ'' = −g/R sin θ, where θ is the angle, R is the length of the rod, and g represents gravitational acceleration. This equation is derived from the rotational analog of Newton's second law, τ = I α, where τ is the net torque and I is the rotational inertia. The discussion confirms that the equation accurately describes the instantaneous angular acceleration of the pendulum based on the angle θ, the length of the rod R, and the gravitational acceleration g.

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StevenJacobs990
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Is this a legitimate equation?
θ'' = − gR sin θ

Source: ftp://www.myphysicslab.com/pendulum1.html
ftp://www.myphysicslab.com/images/pendulum_2.gif
The pendulum is modeled as a point mass at the end of a massless rod. We define the following variables:
  • θ = angle of pendulum (0=vertical)
  • R = length of rod
  • T = tension in rod
  • m = mass of pendulum
  • g = gravitational constant
We will derive the equation of motion for the pendulum using the rotational analog of Newton's second law for motion about a fixed axis, which is τ = I α where
  • τ = net torque
  • I = rotational inertia
  • α = θ''= angular acceleration
The rotational inertia about the pivot is I = m R2. Torque can be calculated as the vector cross product of the position vector and the force. The magnitude of the torque due to gravity works out to beτ = −R m g sin θ. So we have
R m g sin θ = m R2 α

which simplifies to
θ'' = − gR sin θ
 
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Y
StevenJacobs990 said:
Is this a legitimate equation?
θ'' = − gR sin θ
Yes, it is correct. The angular acceleration of the pendulum is ## \ddot {\theta} = -\frac{g\sin(\theta)}{R}##
 
ehild said:
Y

Yes, it is correct. The angular acceleration of the pendulum is ## \ddot {\theta} = -\frac{g\sin(\theta)}{R}##

I have four more questions. I thought making a new forum would be too much for such simple questions.

1. Is "g" gravitational constant or gravitational acceleration? Because in my school I learned lowercase g as gravitational acceleration and capital G as the gravitational constant.

2. What does the equation, θ'' = − gR sin θ, exactly tell?
So if I know g, R, and the angle, then I only know the angular acceleration at that instantaneous moment? Because θ will decrease as pendulum sweeps down.

3. Regarding question 2 that if θ'' changes throughout an oscillation of a pendulum, then is the integral of θ'' verses time graph will be change in angular velocity ω?

4. Regarding question 3 that if I can get angular velocity with integral of angular acceleration v.s. time graph, then it is possible to find the linear velocity when the ball is at its minimum point?
 
StevenJacobs990 said:
I have four more questions. I thought making a new forum would be too much for such simple questions.

1. Is "g" gravitational constant or gravitational acceleration? Because in my school I learned lowercase g as gravitational acceleration and capital G as the gravitational constant.
g is the gravitational acceleration near the surface of the Earth.
StevenJacobs990 said:
2. What does the equation, θ'' = − gR sin θ, exactly tell?
So if I know g, R, and the angle, then I only know the angular acceleration at that instantaneous moment? Because θ will decrease as pendulum sweeps down.
Yes, the formula gives the instantaneous angular acceleration at angle θ.
StevenJacobs990 said:
3. Regarding question 2 that if θ'' changes throughout an oscillation of a pendulum, then is the integral of θ'' verses time graph will be change in angular velocity ω?
Yes, the change of the angular velocity between time instant t1 and t2 is equal to the integral of θ'' with respect time.
StevenJacobs990 said:
4. Regarding question 3 that if I can get angular velocity with integral of angular acceleration v.s. time graph, then it is possible to find the linear velocity when the ball is at its minimum point?
Yes, the linear velocity is v=Rω, at every point of the track.
 

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