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

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Homework Help Overview

The problem involves determining the angular velocity at a specific angle given an angular acceleration that depends on the angular velocity itself. The context is rooted in rotational dynamics, specifically dealing with angular motion and the relationships between angular displacement, velocity, and acceleration.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of angular acceleration with respect to angular displacement and the implications of the limits of integration. There are attempts to reconcile differing results from calculations and the textbook answer.

Discussion Status

Some participants have provided guidance on recalibrating the limits of integration, while others have worked through the calculations step-by-step. There is an acknowledgment of a mistake in the original calculations, but no consensus has been reached on the final value of angular velocity.

Contextual Notes

There are indications of confusion regarding the limits of integration and the relationship between angular velocity and the logarithmic expressions derived during the calculations. The original poster's calculations differ from the expected textbook answer, prompting further discussion.

Bingo1915
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[SOLVED] Angular Velocity

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


2.
Used [tex]\alpha=d\varpi/d\theta * \varpi[/tex]


3.
[tex]\int-2 d\theta=\int1/\varpi d\varpi[/tex]

after integration I got

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

limits 0-60 for [tex]\theta[/tex] and 10 to [tex]\varpifor \varpi[/tex]

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:
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Redo the calculation. Book is correct. The limits of theta should be from pi/6 to pi/3.
 
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:
1/[e^(-pi/3+ln10)] = w

e^(-pi/3+ln10) = w [ln a = b ie e^b = a]
 
Bingo1915 said:
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.)
 
Last edited:
I see where I was incorrect.

Thanks for the help.
 

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