Angular velocity of a rod rotating around a vertical axis

In summary, the conversation discusses how to find the angular velocity needed for a homogeneous rod to rotate around a vertical axis at an angle of theta with the vertical. The equation to calculate this angular velocity is given as \omega = \sqrt{3g/2L cos(θ)}. However, there is some ambiguity in the equation and it is necessary to include extra parentheses to clarify the numerator and denominator. Additionally, there is a mistake in stating that the angular acceleration is equal to d\omega/dθ, when it is actually equal to d\omega/dt.
  • #1
alaix
11
0

Homework Statement



An homogeneous rod is fixed to an extremity and is rotating around a vertical axis, doing an angle of theta with the vertical. (in the scheme I made I wrote alpha but it's theta! (θ))

If the length of the rod is L, show that the angular velocity needed to make it turn is

[itex]\omega[/itex] = [itex]\sqrt{3g/2L cos(θ)}[/itex]
9hCQw.png


Homework Equations



[itex]\tau[/itex] = r x F
I = 1/2 m L²

[itex]\tau[/itex] = I[itex]\alpha[/itex]

The Attempt at a Solution



Here is what I tried

I considered that all the exterior forces (ie. gravity) was acting on the center of mass of the rod, which is situated in the middle, at L/2.

Therefore

Torque = r x F = 1/2 L mg sin(θ)

Torque = I[itex]\alpha[/itex]

Where I = 1/3 mL²

Therefore

[itex]\alpha[/itex] = [itex]\frac{3mgL sin(θ)}{2mL^{2}}[/itex] = [itex]\frac{3g sin(θ)}{2L}[/itex]

Since I'm looking for the angular VELOCITY, and since angular acceleration = d[itex]\omega[/itex]/dθ

[itex]\alpha[/itex] dθ = d[itex]\omega[/itex]

By integrating both sides I find

[itex]\omega = -\frac{3g cos(θ)}{2L}[/itex]


Which is ALMOST the answer I'm looking for... what am I missing?

Thanks!
 

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  • #2
alaix said:

Homework Statement



An homogeneous rod is fixed to an extremity and is rotating around a vertical axis, doing an angle of theta with the vertical. (in the scheme I made I wrote alpha but it's theta! (θ))

If the length of the rod is L, show that the angular velocity needed to make it turn is

[itex]\omega[/itex] = [itex]\sqrt{3g/2L cos(θ)}[/itex]

There is a lot of ambiguity here as to what the final answer is supposed to be. You should include some extra parentheses under the square root sign to make it clear what's in the numerator and what's in the denominator.

alaix said:

Homework Equations



Since I'm looking for the angular VELOCITY, and since angular acceleration = d[itex]\omega[/itex]/dθ

[itex]\alpha[/itex] dθ = d[itex]\omega[/itex]

No, actually that's wrong: [itex] \alpha \neq d\omega/d\theta [/itex]. Rather, [itex] \alpha = d\omega / dt [/itex].
 

1. What is angular velocity?

Angular velocity is a measure of how quickly an object is rotating around a fixed axis. It is often represented by the Greek letter omega (ω) and is measured in radians per second.

2. How is angular velocity calculated?

Angular velocity can be calculated by dividing the change in angular displacement by the change in time. This is represented by the equation ω = Δθ / Δt, where ω is angular velocity, Δθ is change in angular displacement, and Δt is change in time.

3. How does a rod's length affect its angular velocity?

A rod's length does not affect its angular velocity as long as it is rotating around a fixed vertical axis. The angular velocity is only affected by the object's mass distribution and the force acting on it.

4. Can angular velocity change over time?

Yes, angular velocity can change over time if the object's speed of rotation changes or if the axis of rotation changes. This change in angular velocity is known as angular acceleration.

5. How is angular velocity different from linear velocity?

Angular velocity measures the rate of change of an object's rotation, while linear velocity measures the rate of change of an object's linear motion. They are related by the formula v = rω, where v is linear velocity, r is the distance from the axis of rotation, and ω is angular velocity.

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