# Centripetal acceleration units

I perform some basic calculation for circular motion. Formulas we get from school are:

• $v$ - Linear speed with the units m/s
• $r$ - Radius of curve in meters
• $\omega$ - Angular speed with the units rps (radians per second)
• $a_c$ - Centripetal acceleration with the units m/s${}^2$

$$\omega = \frac{v}{r},$$

$$v = \omega r,$$

$$a_c = \frac{v^2}{r} = \omega^2 r.$$

So far it looks straight forward. My problem is in units.

The units of $\omega^2 r$ are $\dfrac{rad\cdot rad\cdot metre}{sec\cdot sec}$. Acceleration units must be $\dfrac{metre}{sec\cdot sec}$.

• Radian is now considered an SI derived unit (dimensionless). Commented Dec 31, 2014 at 11:42
• More on radians: physics.stackexchange.com/q/33542/2451 and links therein. Commented Dec 31, 2014 at 12:32

Radian $(\theta)$ is defined as, $\theta=\dfrac{l}{r}$, where $l$ is length of arc and $r$ is radius in a circle, and both have dimension as lengths.

Thus, Radian is a dimensionless unit.

The answer to this question has essentially been given by @DheerajKumar, but let's explain it in a different way.

You are studying circular motion, that is the motion of a particle along a circular path. In such motion, linear speed $v$ gives you the length of arc covered by the particle per unit of time. With angular speed $\omega$ you want to give a measure of the angle covered by the particle per unit of time.

But, how to define angle? There is an easy way to do it if you notice that the length of arc of any circular path is proportional to the radius $r$ of the circle. Thus you can define the angle covered by the particle as the length of arc covered by it divided by the radius $r$ of the circle. As stated by @DheerajKumar, this is the definition of radian, that is, if you are defining angle in that way, you are measuring it in a unit called radian. With that definition of angle you obtain

$$\omega = \frac{v}{r},$$

which is equivalent to

$$v = \omega \cdot r.$$

Notice that these formulas are only valid if you use radians as the unit of angle.

To understand the units, notice that $\rm radians$ are $\;\dfrac{meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)}}{meters\ {\rm \scriptstyle (coming\ from\ radius)}}$. The units of $a_c = \dfrac{v^2}{r}$ are $$\frac{ meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)} \cdot meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)} } { meters\ {\rm \scriptstyle (coming\ from\ radius)} \cdot sec \cdot sec } = {\rm m/s^2}.$$ You can see is the same thing if you use $a_c = \omega^2 r$ which has the units $$\frac {rad\cdot rad\cdot meters\ {\rm \scriptstyle (coming\ from\ radius)} } {sec\cdot sec} = \frac{ meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)} \cdot meters\ {\rm \scriptstyle (coming\ from\ length\ of\ arc)} \cdot meters\ {\rm \scriptstyle (coming\ from\ radius)} } { meters\ {\rm \scriptstyle (coming\ from\ radius)} \cdot meters\ {\rm \scriptstyle (coming\ from\ radius)} \cdot sec \cdot sec } = {\rm m/s^2}.$$

Consider two lines intersecting and making some angle. If you draw any circle non-zero radius, whose centre lies at the point of intersection of the two lines, then the length of arc cut off by the lines on the circumference of the circle, divided by its radius is defined as the angle between the lines in radians.

As you can see, radians will have units of meter/meter ( They have no units). So radians are just numbers, having no unit. Even in angular velocity, the technically correct unit is 1/second.