Angular acceleration in terms of angular velocity

In summary, the book explains how to find the angular frequency of a physical pendulum using the SHM equation of motion. The equation is derived using the small-angle approximation and the angular frequency is found to be a cosine with √k/m.
  • #1
Flipmeister
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I've been working on problems that deal with pendulums and I've noticed that a few of my answers require me to find the angular velocity, frequency, period of a pendulum. I managed to get the answer right every time, but there's a step that I didn't understand, namely converting angular acceleration to angular velocity by what seems to be using this relationship:

[itex]\alpha=\omega^2[/itex]

Where in the heck does this come from? :confused: The book kind of skips this step in the examples and doesn't prove it at all, even though it proves just about every other equation it mentions.
 
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  • #2
That looks like a conversion factor for radial acceleration.

F=m * v2/r = m * a(rad)

so a(rad) is proportional to w2.

You'd need to show the example in detail for anyone to say more than that.
 
  • #3
Here's how the book derives ##w=\sqrt{\frac{Mgl}{I}}##, the angular frequency of a physical pendulum:

[tex]\tau=-Mgd=Mglsin\theta[/tex]

If we restrict the angle to being small (<10 degrees), as we did for the simple pendulum, we can use the small-angle approximation to write

[tex]\tau=-Mgl\theta[/tex] (eq 14.49)

From Chapter 12, Newton's second law for rotational motion is ##\alpha=\frac{d^2\theta}{dt^2}=\frac{\tau}{I}## where I is the object's moment of inertia about the pivot point. Using eq. 14.49, we find

[tex]\frac{d^2\theta}{dt^2}=\frac{-Mgl}{I}\theta[/tex]

Comparison with eq. 14.32 (##\frac{d^2x}{dt^2}=-\frac{k}{m}x##) shows that this is again the SHM equation of motion, this time with angular frequency

[tex]\omega=2\pi f=\sqrt\frac{Mgl}{I}[/tex]

Something happens to get to that last step that doesn't add up for me... Why is Mgl/I square rooted?
 
  • #4
It's justified here by comparison to an earlier expression for SHM Eq 14:32. In that section you will probably find a proof that f=1/2*pi sqrt(k/m)

There's a fair explanation in http://en.wikipedia.org/wiki/Simple_harmonic_motion

It's just that the solution of that differential equation is a cosine with √k/m (you can prove that to yourself by substituting it back)
 
  • #5


Hello,

Thank you for sharing your observations and questions about angular acceleration and angular velocity in relation to pendulums. The relationship between angular acceleration and angular velocity is a fundamental concept in rotational motion and it is crucial in understanding the behavior of pendulums.

To address your question about the relationship \alpha=\omega^2, this comes from the definition of angular acceleration and the definition of angular velocity. Angular acceleration is defined as the rate of change of angular velocity, and angular velocity is defined as the rate of change of angular displacement. Therefore, we can write the following equations:

\alpha=\frac{d\omega}{dt}

\omega=\frac{d\theta}{dt}

Where \alpha is angular acceleration, \omega is angular velocity, and \theta is angular displacement. If we substitute the second equation into the first one, we get:

\alpha=\frac{d}{dt}\left(\frac{d\theta}{dt}\right)

Using the chain rule, we can rewrite this as:

\alpha=\frac{d^2\theta}{dt^2}

This is the familiar equation for angular acceleration in terms of angular displacement. However, for a pendulum, the angular displacement \theta is related to the angular velocity \omega by the formula \theta=\omega t. Substituting this into the above equation, we get:

\alpha=\frac{d^2}{dt^2}(\omega t)

Using the product rule, we can rewrite this as:

\alpha=\omega\frac{d^2t}{dt^2}+t\frac{d^2\omega}{dt^2}

Since \frac{d^2t}{dt^2}=0 (since time is not changing), we can simplify this to:

\alpha=\omega\frac{d^2\omega}{dt^2}

Finally, using the definition of angular acceleration, we can write this as:

\alpha=\omega\frac{d\omega}{dt}

And since we know that \alpha=\omega^2, we can rearrange this equation to get:

\omega^2=\omega\frac{d\omega}{dt}

This is the relationship you mentioned, and it comes from the fundamental definitions of angular acceleration and angular velocity.

I hope this explanation helps to clarify the concept for you. It is important to understand the relationships between different quantities in physics, as they can help us solve problems and make predictions about the
 

What is angular acceleration in terms of angular velocity?

Angular acceleration is a measure of the rate at which an object's angular velocity changes over time. It is usually denoted by the symbol alpha (α) and is expressed in units of radians per second squared (rad/s^2).

How is angular acceleration related to linear acceleration?

Angular acceleration is related to linear acceleration through the radius of rotation. Linear acceleration is equal to the product of angular acceleration and the radius of rotation. This relationship is described by the formula a = αr, where a is linear acceleration, α is angular acceleration, and r is the radius of rotation.

What is the difference between angular acceleration and angular velocity?

Angular acceleration and angular velocity are both measures of rotational motion, but they are different quantities. Angular velocity is a measure of how fast an object is rotating, while angular acceleration is a measure of how quickly the angular velocity is changing.

How is angular acceleration calculated?

Angular acceleration can be calculated by dividing the change in angular velocity by the change in time. The formula for angular acceleration is α = (ω2 - ω1) / (t2 - t1), where α is angular acceleration, ω is angular velocity, and t is time.

What factors can affect the angular acceleration of an object?

The angular acceleration of an object can be affected by factors such as the applied torque, the moment of inertia, and any external forces acting on the object. The shape and mass distribution of an object can also affect its angular acceleration.

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