- #1
gangsta316
- 30
- 0
Is this derivation correct?
I managed to derive [itex] v = r \omega[/itex] and (I think) [itex] a = \omega^2 r[/itex].
I did [itex] a^2 = \ddot{x}^2 + \ddot{y}^2[/itex] to get eventually
[itex]a^2 = r^2(\dot{\theta}^4 + \ddot{\theta}^2)[/itex]. Then I said that, for uniform circular motion, the angular velocity [itex]\dot{\theta}[/itex] is a constant [itex]\omega[/itex] meaning that [itex]\ddot{\theta}[/itex] is equal to zero. So then it came to [itex] a = \omega^2 r[/itex].
But how do we know, from my derivation, that this is a centripetal acceleration toward the center? Is it because the tangential acceleration is the one I discarded and so this is the perpendicular component of acceleration? Is the formula valid where the angular velocity is not constant?
I managed to derive [itex] v = r \omega[/itex] and (I think) [itex] a = \omega^2 r[/itex].
I did [itex] a^2 = \ddot{x}^2 + \ddot{y}^2[/itex] to get eventually
[itex]a^2 = r^2(\dot{\theta}^4 + \ddot{\theta}^2)[/itex]. Then I said that, for uniform circular motion, the angular velocity [itex]\dot{\theta}[/itex] is a constant [itex]\omega[/itex] meaning that [itex]\ddot{\theta}[/itex] is equal to zero. So then it came to [itex] a = \omega^2 r[/itex].
But how do we know, from my derivation, that this is a centripetal acceleration toward the center? Is it because the tangential acceleration is the one I discarded and so this is the perpendicular component of acceleration? Is the formula valid where the angular velocity is not constant?