Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let's consider a cicumference that have the center in the origin of axes and rotates around x-axes. Let's stick a bar in a point $A$ of this circumference and at the end of the bar let's stick a mass point $P$. Let's call $\theta$ the angle that the radius $OA$ formes with z-axes and $\phi$ the angle that $AP$ formes with z-axes. How can I find the angular velocity of $P$?

At the beginnig I have thought that angular velocity must be the same for all the points inner and on the circumference. And so I have thought that it could be $\dot \theta$. The book gives me this result for the kinetic energy of P: $\frac{1}{2}m(\dot \theta^2+\dot \phi^2+2\dot \theta \dot \phi cos (\theta-\phi))$.

Considering that in this case the kinetic energy is only rotational and that general expression of kinetic rotational energy for a mass point is $\frac {1}{2}mr^2\omega^2$ I can't understand the origin of this result... and I can't understand why I have to consider the two angles. Could you help me?

share|improve this question
1  
Sure, that's better. In particular, your third paragraph where you say you can't understand the origin of the result $\frac{1}{2}m(\dot\theta^2 + \cdots)$ is great. Asking what you have to do (as in the last paragraph), not so much. –  David Z Jan 5 '13 at 20:21
    
@DavidZaslavsky Is it right now? –  sunrise Jan 5 '13 at 20:58
    
Yeah, this is okay. :-) –  David Z Jan 5 '13 at 21:09
    
Hint, Check the kinetic energy units. There must be a length$^2$ quantity in there for it to work out, and hence the kinetic energy is not purely rotational, but has translational components. –  ja72 Aug 12 '13 at 16:22
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.