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David DanaBashian
Joined: 28 May 2008 Posts: 3 Location: ddb4chess


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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Sat Jun 07, 2008 12:40 am Post subject: 


I think it is correct. The analogy is between linear motion and rotational motion. The formula for linear motion is in terms of velocity. So the rotational analogue is in terms of the angular velocity, omega. 

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David DanaBashian
Joined: 28 May 2008 Posts: 3 Location: ddb4chess

Posted: Sat Jun 07, 2008 1:34 am Post subject: 


I disagree. Drop theta0 and let theta correspond to (xx0) if you like, but otherwise, my angular formula is correct and your third angular formula is incorrect.
In your other three pairs of equations, omega corresponds to v, omega0 corresponds to v0, alpha corresponds to a, etc. In your third angular formula, you have omega corresponding to x while omega0 still corresponds to v0.
Do you still need more proof that your third angular formula is incorrect? Since omega0*t and (1/2)*alpha*t*t on the righthand side of the equation are both distances, so is their sum. Therefore the term on the lefthand side of the equation must be a distance, but it is omega, an angular velocity, instead!
Do you still need more proof that your third angular formula is incorrect? Compare your first angular formula to your third angular formula. In general, both these formulae, as you have stated them, cannot be true at the same time! 

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hepcj Site Admin
Joined: 23 Jun 2007 Posts: 125

Posted: Sun Jun 08, 2008 12:03 am Post subject: 


You are right it. It is a typo. Thanks. 

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David DanaBashian
Joined: 28 May 2008 Posts: 3 Location: ddb4chess

Posted: Sun Jun 08, 2008 1:39 am Post subject: angular velocity and angular acceleration 


Now you must decide whether to let theta stand for (xx0), or to let theta stand for x and theta0 stand for x0. Right now your equations display a mixture of the two replacements. Please decide on one replacement and correct accordingly. 

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