What is the Angular Velocity at a Given Angle for a Given Angular Acceleration?

seboastien
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Homework Statement



angular acceleration of line a= -2w^2 rad/s

when theta=pi/6, w= 10 rad/s

what is angular velocity when theta= pi/3?

Homework Equations



chain rule

The Attempt at a Solution



w= e^-2theta is my answer, but I know this is wrong

I know that this question is really easy, but I just don't know how to get it
 
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Well you should know that α= -2ω2 means that dω/dt = -2ω2

which is an easily solvable DE once you 'separate the variables' i.e. put all the things with ω on one side and t on the other then integrate both sides.
 
But then I end up with t=1/2w, how do I find the rest of the answer?
 
Someone please help me!
 
seboastien said:
But then I end up with t=1/2w, how do I find the rest of the answer?

I now saw that you have ω and θ so forget my above post and replace α with ω*dω/dθ and integrate again. Do not forget the constant of integration which you can find with the conditions they gave you.
 
Yes, that was my problem all along... I forgot to add the constant! Thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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