# Normal and tangential components of acceleration

I was asked to solve the following problem on a test:

A motorist is moving at 70 km/hour on a circular path of radius 500 meter. Determine the normal and tangential components of acceleration.

All I know from high school physics knowledge - centripetal acceleration in uniform circular motion is $\frac{v^2}{r}$. I am not sure whether this is the normal or tangential component of acceleration or none. If this is one of those two, then how to calculate the other one? What are ways to distinguish them?

If the speed is constant there is only normal acceleration (from the motorist to the center of the circunference).

To get both accelerations you can make:

1) Get your position vector in polar coordinates: $$\vec{r} = x\vec{i}+y\vec{j} =r \cos \theta \; \hat{i} + r \sin \theta \; \hat{j} = r \hat{r}.$$

Now you can get the speed vector making the derivative with respect to time: $$\vec{v} = \dfrac{d\vec{r}}{dt}=\omega r \left( - \sin \theta \; \hat{i} + \cos \theta \; \hat{j}\right) = \omega r \hat{v},$$ wehre $$\omega = \dfrac{d\theta}{dt}.$$

Finally you get the acceleration $$\vec{a}=\dfrac{d\vec{v}}{dt}= \omega^2 r \left( -\cos \theta \; \hat{i} - \sin \theta \; \hat{j} \right) + r\alpha \left( - \sin \theta \; \hat{i} + \cos \theta \; \hat{j}\right),$$ where $$\alpha = \dfrac{d\omega}{dt}.$$

Now notice that you can rewrite the acceleration in terms of the position and speed vectors: $$\vec{a}= -\omega^2 r \hat{r} + r\alpha \hat{v},$$ and that's it. One of it is the normal acceleration and the other is the tangential. Who to rembember it? Easy: position vector goes from the center of the circle, so $-\omega^2 r \hat{r}$ is an acceleration that goes to the center of the circle: we call it normal or radial acceleration $a_n=\omega ^2 r$. The other acceleration goes in the speed vector, and the speed is alway tangential to the circle in circular motion. Then we call ir tangential acceleration $a_t = \alpha r$.

If you go back to the original coordinates, you get what you said: $$a_n = \dfrac{v^2}{r}, \qquad a_t = \dfrac{dv}{dt}.$$ In your homework notice that you hace a 70 km/h speed with no acceleration. Then one of the circular accelerations is zero. Guess what? Notice also that in a circular motion theres is ALWAYS one acceleration that MUST be non-zero.

• So please give me a positive point to the answer – Carlos Oct 29 '17 at 15:26
• Sure, but how to do that? – hellomecha Oct 29 '17 at 16:25
• With the arrows in the left of the message. Yo have an up arrow then a zero and then down arrow – Carlos Oct 29 '17 at 21:10

In uniform circular motion the acceleration is radial (normal) of modulus v^2/r. The tangential component is zero.

Tangential acceleration is acceleration parallel to the circular path. Since the velocity (this velocity is also in tangential direction) of the motorist is constant, there is no tangential acceleration.

Normal acceleration is acceleration perpendicular to the path; here it is the centripetal acceleration.