Determine the angular velocity as a function of the angle

[haruspex]

Ok, So it would be $\frac{\omega^2}{2} = \beta_0 \sin\varphi+0$, right? I understand, thanks! How do I draw the graph of the dependence?

It just means sketch a graph of ω against φ.

 

[Davidllerenav]

It just means sketch a graph of ω against φ.

But how to I graph that?

 

[haruspex]

But how to I graph that?

How do you sketch any graph?
Pick a few convenient values of x and see what y is.
Same for gradients at the same points.
Consider local extrema and asymptotic behaviour.

 

[Davidllerenav]

How do you sketch any graph?
Pick a few convenient values of x and see what y is.
Same for gradients at the same points.
Consider local extrema and asymptotic behaviour.

But there isn't any x or y. It is polar coordinates, right?

 

[haruspex]

But there isn't any x or y. It is polar coordinates, right?

You are just sketching a graph of one variable against another. The physical meanings are irrelevant.
You could sketch it in polar if you wish, but I doubt that is what is expected.

 

[Davidllerenav]

You are just sketching a graph of one variable against another. The physical meanings are irrelevant.
You could sketch it in polar if you wish, but I doubt that is what is expected.

So I have to set $\omega$ as y and $\varphi$ as x, right? Then I you said I just need to plug in values. The values that would be easy to cumpute would be $\varphi=0$, $\varphi=1$, $\varphi=\pi$, $\varphi=\pi/2$, right?

 

[SammyS]

But there isn't any x or y. It is polar coordinates, right?

To plot angular velocity, $\omega ,$ as a function of $\varphi$, the usual convention is to plot $\varphi$ on the x-axis and angular velocity, $\omega ,$ on the y-axis.

 

[Davidllerenav]

To plot angular velocity, $\omega ,$ as a function of $\varphi$, the usual convention is to plot $\varphi$ on the x-axis and angular velocity, $\omega ,$ on the y-axis.

Like I said on post #36?

 

[SammyS]

Like I said on post #36?

Well, yes for:

So I have to set $\omega$ as y and $\varphi$ as x, right?

However, the following isn't quite right.

The values that would be easy to cumpute would be $\varphi=0$, $\varphi=1$, $\varphi=\pi$, $\varphi=\pi/2$, right?

At least, $\varphi=1$ is not an easy value, is it?

Other values "easy" for $\varphi$ are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$.

 

[Davidllerenav]

At least, $\varphi=1$ is not an easy value, is it?

Oh, yes it isn't, sorry.

Other values "easy" for φφ\varphi are π6π6\dfrac{\pi}{6} and 5π65π6\dfrac{5\pi}{6} .

Yes, you're right.