0

We have this formula for centripetal acceleration - $a = v \frac{d\theta}{dt} = v\omega = \frac{v^2}{r}$

but in case of usual acceleration i know that speed in $t_1 = v_0 + a \cdot (t_1 - t_0)$

but in circular case i don't understande the nature of acceleration, speed is always the same if motion is uniform, but acceleration is not zero.

EDIT: Suppose T is 1. $t_0 = 0, t_1 = .25$, so $v$ was rotated for $\frac{\pi}{2}$. So $\frac{\bar{a}}{4} = \bar{v_1} - \bar{v_0}$ and $|a| = 4\sqrt{v^2 + v^2} = 4\sqrt{2} v$, but with previous formula magnitude of a is $\frac{v^2}{4r}$ .

ANSWER: physical meaning of $a$ is the length of an arc swept out by velocity vector!

Yola
  • 310
  • 1
    Suppose you have a point moving along a circle. The point's distance from the center never changes, yet it's velocity is non-zero. What gives? – David H Dec 31 '13 at 13:55
  • 2
    The magnitude of $\vec{v}$ is constant, but not it's direction. – jinawee Dec 31 '13 at 14:04
  • There is much wrong with your question - namely the units don't work out for most of your equations except for the very first one. It's hard then to decipher what exactly you're asking. Jinawee makes a good point that you should keep in mind: Acceleration is defined by the $\frac{d \vec{v}}{dt}$ and since $\vec{v}$ is a vector, then if the direction of $\vec{v}$ changes in time (even with it's magnitude constant) then there is a non-zero acceleration. – mcFreid Dec 31 '13 at 15:00
  • mcFreid, vector - vector is vector, magnitudes have no units at all. – Yola Dec 31 '13 at 15:37
  • 1
    See http://physics.stackexchange.com/q/38291/ and http://physics.stackexchange.com/q/91797/ – John Rennie Dec 31 '13 at 15:55
  • 2
    I have no idea what you mean by "vector is vector." Magnitudes can have units. For example, the magnitude of the velocity is the speed and it has units of $\frac{m}{s}$. – mcFreid Dec 31 '13 at 16:37
  • The analysis in your edit is flawed in a couple of ways. (1) $T$ is not independent, but is a function of $v$ and $r$ so you should get $\bar{a} = (\sqrt{2} v)/(0.25 \cdot 2 \pi r / v) = (2\sqrt{2} v^2) / (\pi r)$. (2) The result depends on how much of a rotation you use, doing the same thing for a half rotation yields $\bar{a} = (2 v) / (0.5 \cdot 2 \pi r / v) = (2 v^2)/(\pi r)$ which is not the same as for a quarter rotation demonstrating that the analysis is not correct (I suspect tat you can repair it by taking the infinitesimal limit). – dmckee --- ex-moderator kitten Dec 31 '13 at 17:44
  • You wrote: $|a| = 4\sqrt{v^2 + v^2} = 4\sqrt{2} v$, it should be $|\vec{a}| = 4\sqrt{v_x^2 + v_y^2} = 4v$. Anyway, $\frac{\bar{a}}{4} = \bar{v_1} - \bar{v_0}$ is clearly wrong. – jinawee Dec 31 '13 at 17:46

1 Answers1

0

Physical meaning of $a$ is the length of an arc swept out by velocity vector.

Yola
  • 310