Solve Angular Velocity: \theta=60deg, \alpha=-2w^2

[Bingo1915]

[SOLVED] Angular Velocity

1.
The angular acceleration is given $$\alpha$$=-2w^2 rad/s^2 where $$\omega$$ is the angular velocity in rad/s. When $$\theta$$=30 deg the angular velocity is 10 rad/s. What is the angular velocity when $$\theta$$=60deg?


2.
Used $$\alpha=d\varpi/d\theta * \varpi$$


3.
$$\int-2 d\theta=\int1/\varpi d\varpi$$

after integration I got

[$$\(-2)*theta$$]=[$$ln\varpi$$]

limits 0-60 for $$\theta$$ and 10 to $$\varpifor \varpi$$

I think I'm missing a step somewhere. The book gives an answer of w=3.51. With my calculations I get w=1.14. Can you advise?

 

[Shooting Star]

Redo the calculation. Book is correct. The limits of theta should be from pi/6 to pi/3.

 

[Bingo1915]

Using pi/6 to pi/3 I get

-2pi/3 + pi/3 = lnw-ln10

-pi/3 + ln10 = lnw

From here I think it is

1/[e^(-pi/3+ln10)] = w

Can you check me on this part?

 

[Leong]

1/[e^(-pi/3+ln10)] = w

e^(-pi/3+ln10) = w [ln a = b ie e^b = a]

 

[Shooting Star]

Using pi/6 to pi/3 I get

-2pi/3 + pi/3 = lnw-ln10

-pi/3 + ln10 = lnw

ln 10/w = pi/3. Take antilog of pi/3, use a calulator (or something) => w = 3.51.

(By antilog, I meant, 10/w = e^pi/3. You have done everything correctly.)

 

[Bingo1915]

I see where I was incorrect.

Thanks for the help.