# Angular Velocity

1. Jan 20, 2008

### 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?

Last edited: Jan 20, 2008
2. Jan 21, 2008

### Shooting Star

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

3. Jan 21, 2008

### 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?

Last edited: Jan 21, 2008
4. Jan 22, 2008

### Leong

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

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

5. Jan 22, 2008

### Shooting Star

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.)

Last edited: Jan 22, 2008
6. Jan 22, 2008

### Bingo1915

I see where I was incorrect.

Thanks for the help.