I was doing a physics problem and realized that the formula for angular acceleration was the same as the formula for centripetal acceleration (in terms of angular speed) They both are [tex]\omega^2r[/tex] where w is angular speed and r is the radius Why is that so? When I tried to derive this I got... [tex]a_{centripetal}=\frac{v^2}{r}[/tex] [tex]a_{centripetal}=\frac{v}{r}v[/tex] since v/r=w then... [tex]a_{centripetal}=\omega v[/tex] how are they equal? [tex]\omega v=\omega^2r[/tex]
You've got it wrong ,check your book true centripetal acceleration is v^2 /r BUT ANGULAR acceleration is the something spinning up faster or spinning down slower. Angular acceleration is [tex] \alpha = \frac{\Delta \omega}{\Delta t} = \frac{v}{rt} [/tex]
Here is the question...http://home.earthlink.net/~suburban-xrisis/physics001.jpg the answer is A. why?
[tex] \omega [/tex] is the ANGULAR VELOCITY (or frequency) and [tex] \omega = \frac{v}{r} [/tex] the CENTRIPETAL ACCELERATION is a = v^2 /r since omega = v/r a = omega r
You have to work out the change in velocity as a function of its tangential speed, [itex]v [/itex] or [itex] \omega r[/itex]. Draw a diagram of the velocity vector of a unit mass at time 0. Then draw its velocity vector after a time dt. The mass turns through an angle [itex]d\theta = ds/r = \frac{vdt}{r}[/itex] in that time. Also remember that [itex]v = 2\pi r/T = \omega r[/itex] and [itex]d\theta = \omega dt[/itex] Now, the new velocity vector at t=dt is the same length as at t=0 but pointed [itex]d\theta[/itex] to the original. The difference is the change in velocity or dv and is directed toward the centre of the circle along the radius. You can see from the diagram that: [itex] dv = vsin(d\theta)[/itex] which approaches the limit of [itex] dv = vd\theta[/itex] as [itex]d\theta \rightarrow 0[/itex]. This means: [tex]dv = vd\theta = \omega r d\theta = \omega^2r dt[/tex] so [tex]dv/dt = a_{centripetal} = \omega^2r[/tex] AM
what is wrong with my method? I subbed in v/r for omega but got r*omega I understand that if I subbed v=omega*r then the equation would come out correct
No. They are two distinct concepts; two quite different vector quantities with different directions. For a mass moving in a curved path, centripetal acceleration is radial - toward the centre. Tangential acceleration - in the direction of travel gives rise to non-zero angular acceleration. The centripetal acceleration ([itex]a_c = -\omega^2r[/itex]) is always non-zero if there is circular motion. AM
yes I understand the concepts are different, but both equations can be expressed as [tex]\omega^2 * r[/tex] is that correct?
No. Angular acceleration has nothing to do with [itex]\omega[/itex]. It depends on torque not angular speed, just as acceleration is a function of force not velocity. The definition of angular accelaration is [itex]\alpha = a/r = f/mr = fr/mr^2 = \tau/mr^2[/itex]. So [itex]\tau = m\alpha r^2[/itex] AM