Angular frequency of a physical pendulum

[Pushoam]

Homework Statement

For calculating angular frequency of a physical pendulum, I consider its center of mass motion.
The COM motion is a simple pendulum motion.
Considering a coordinate system whose origin is the pivot point. Then, the COM is the length of the corresponding simple pendulum. Is it correct to say that the angular frequency of this simple pendulum is the angular frequency of the physical pendulum?

Homework Equations

The Attempt at a Solution

 

[jambaugh]

The distance from the COM to the pivot is a good approximation but is not the true effective length of a physical pendulum.

The true quantity one is looking for is the moment of inertia about the pivot. For a point mass that will be the length times the mass. For an extensive physical object there will be an additional contribution from the moment of Inertia of the object about its own COM. This will add to the effective length since both more length and a object with higher moment of inertia will contribute to the total moment of inertia.

I worked out the details but if this involves a homework problem you do as much work as you can yourself before looking at my exposition.

Spoiler

Let me see here... You'll have a moment of inertia: $I = \ell^2 M + I_0$ where $I_0$ is the object's moment of inertia when rotated about its own COM.

The torque on the pendulum at a given by the force of gravity times its lateral displacement which, in terms of the angle from downward, is $\tau = - gM \ell \sin(\theta)\approx -gm\ell \theta$. Here $\ell$ is the distance from the COM to the pivot and we're using the small angle approximation (keep the maximum swing under $10^\circ$).
The angular acceleration will relate to the torque by $I \ddot{\theta} = \tau$ so $\ddot{\theta} = -\frac{gM\ell}{M\ell^2+I_0} \theta$
In terms of the angular frequency of the pendulum $\ddot{\theta} = -\omega^2 \theta$.
This gives an angular frequency of $\omega^2 = g\frac{M\ell}{M\ell^2 + I_0}$ We typically express moments of inertia for objects about their center in the form of a constant times their mass. So let the moment of inertia here be $I_0 = kM$. Then a little algebra gives us:
$$\omega = \sqrt{ \frac{g}{\ell + k/\ell}}$$.The effective length is $\tilde{\ell} = \ell + \frac{k}{\ell}$ where $k = I_0/M$ is the ratio of the objects central moment of inertia over its mass.
For example if the object is a solid ball of radius $a$ the moment of inertia is $\frac{2}{5} Ma^2$ and so $k = 2a^2/5$

Then the effective length of a pendulum using that ball weight is: $\tilde{\ell} = \ell + \frac{2a^2}{5\ell}$.

 

[Pushoam]

Let's consider a rod of length 'l'.
The torque about P : $\vec \tau =\frac { mgl \sin \theta ~(- \hat z)} 2$
Using work kinetic energy theorem,
work done by this torque = change in the rotational kinetic enerrgy
taking $\theta_i = \pi /2$ and initial rotational kinetic energy to be 0,
$mgl\cos \theta = I \omega ^2 \\ \omega = \sqrt{ ( \frac {mgl\cos \theta} I )}$
This means that the angular velocity depends on the angle theta.
Now the angular velocity is known as angular frequency ,right?
Is this correct so far?
I haven't seen the spoiler till now. I thought I will see it after solving it myself.

Considering COM motion,
loss in the potential energy of the COM = gain in the kinetic energy of the COM
$\frac 1 2 mgl(1-\cos\theta) = \frac 1 2 mv^2 = \frac 1 2 m \omega^2 R^2$ where R is the COM from the pivot
$\omega = \sqrt {\frac {gl(1-\cos\theta)}{R^2}}$

Here the $\omega$'s turn out to be different.
Is something wrong here?

 

[jambaugh]

Check your angular frequency formula. It should not contain the angle. It derives from the form of a harmonic oscillator where the variable's acceleration is negative some constant times its position $\ddot{x} = -kx$. The constant is $\omega^2$. Here you'll have $\ddot{\theta} = -k \sin(\theta)$ and use the small angle approximation $\sin(\theta)\approx \theta$.

 

[Pushoam]

Check your angular frequency formula. It should not contain the angle. It derives from the form of a harmonic oscillator where the variable's acceleration is negative some constant times its position $\ddot{x} = -kx$. The constant is $\omega^2$. Here you'll have $\ddot{\theta} = -k \sin(\theta)$ and use the small angle approximation $\sin(\theta)\approx \theta$.

Does this mean that whenever I have to calculate angular frequency, I have to use the force equation or torque eqn.?

Are angular frequency and angular speed not always same?
Angular velocity calculated using work energy theorem is not equal to angular frequency. Right?
I want to know why are the angular velocity calculated by the two methods different?
They should be same. Right?

 

[jambaugh]

To answer your questions let me explain about the two different angles involved with the simple pendulum.
The pendulum is analogous to a mass spring system because (for the small angle approximation) the generalized force is proportional to the negative of the displacement, and thus too is the acceleration.

For a mass spring system: $\ddot{x} = -\frac{k}{m}x -\omega^2 x$
The $\omega$ here is the angular frequency. Now with the mass and spring there is no geometric angle involved. The angle to which this refers is the angle in phase space where we graph position vs momentum or position vs velocity. In that frame we will see the system state trace out an ellipse parameterized by:
$x = x_\max \cos(\omega t)$, $v = v_\max\sin(\omega t)$ (and we can add a phase angle to indicate different initial configurations and we can let $\omega$ be negative to indicate a time reversed case.)
The angle here is an abstract phase angle and it is the rate of change of this that (in radians per time unit) is referred to as the angular frequency. The oscillation frequency will thus be the angular frequency divided by $2\pi$ since there is 1 cycle per $2\pi$ radians.

Now the pendulum has its own geometric angle defining its geometric configuration. This angle is the system coordinate and its coordinate velocity will thus be an angular velocity with magnitude defining its angular speed. The only sense in that this angular speed would translate into an angular frequency is if you at that given instant remove the restoring force from the pendulum and it became a mass spinning about a pivot. The angular frequency of that spin will be the instantaneous angular speed of the pendulum at the point you magically removed gravity.

So in short, "No" angular speed is distinct from angular frequency for a pendulum and one critical point is that angular frequency is fixed by the pendulum length while angular speed will vary throughout the swing of the pendulum and is the generalized coordinate speed for this general harmonic oscillator.

 

[Pushoam]

Thanks for the clarification.