Determine the angular velocity as a function of the angle

In summary: Ok, so I'll write it like ##\frac{\omega^2}{2} = \beta_0 \sin\varphi+C##. Now, how do I determine its value? Since I don't have the initial conditions, would it be something like...... ##\frac{\varphi_0}{2}##?In summary, the angular velocity of a solid body rotating around a fixed axis is proportional to the cosine of the angle of rotation.
  • #36
haruspex said:
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?
 
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  • #37
Davidllerenav said:
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.
 
  • #38
SammyS said:
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?
 
  • #39
Davidllerenav said:
Like I said on post #36?
Well, yes for:
Davidllerenav said:
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} ##.
 
  • #40
SammyS said:
At least, ## \varphi=1## is not an easy value, is it?
Oh, yes it isn't, sorry.
SammyS said:
Other values "easy" for φφ\varphi are π6π6\dfrac{\pi}{6} and 5π65π6\dfrac{5\pi}{6} .
Yes, you're right.
 
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