# What is the real difference between radial and tangential acceleration?

So in my physics coursebook there are two different kinds of derivation of $$\frac{dv}{dt}$$ of a particle rotating in a circle. Most of you will know these, they are what is called centripetal/radial acceleration $$\frac{v^2}{R}$$ and tangential acceleration $$R\alpha$$. Suppose $$R$$ is radius, $$l$$ is arc-length, $$\theta$$ is angular position in radians, and $$\omega$$ is angular velocity, then the first derivation is something like $$a_{rad}=\frac{dv}{dt}=\frac{v\,d\theta}{dt}=\frac{v\,dl}{R\,dt}=\frac{v^2}{R}\,(=R\omega^2),$$ and the second $$a_{tan}=\frac{dv}{dt}=R\frac{d\omega}{dt}=R\alpha.$$ Now that's clear and all until you start asking questions about how starting from the same origin and unpacking the same derivation actually gives you two very different results.

So I was trying to figure this out, and I thought that maybe it would clear up if I tried doing this with vector quantities, as then you can assume in the tangent case that the unit velocity-vector $$\vec v_u$$ is constant, and in the radial case that speed $$v$$ is constant, and variable the other way around. Then $$\vec a_{tan}$$ is easy: $$\vec a_{tan}=\frac{dv}{dt}\vec v_u=R\frac{d\omega}{dt}\vec v_u=R\alpha\,\vec v_u,$$ but I get stuck with the radial case: $$\vec a_{rad}=\frac{d\vec v}{dt}=\frac{v\,d\vec v_u}{dt}=R\omega\frac{\,d\vec v_u}{dt}=\ldots$$ That's about where I'm at. I've looked for other answers but didn't find the one.

• the direction? (answer below) Nov 2, 2022 at 10:34

From the question, $$\vec a=\frac{d\vec v}{dt}=\frac{d\vec{\omega}}{dt}\times \vec R+\vec{\omega}\times \frac{d\vec R}{dt}=\vec a_t+\vec a_r$$ where, $$\vec v=\vec{\omega}\times \vec R$$ and $$d\vec R$$ is perpendicular to $$\vec R$$, so direction of $$\vec a_t$$ is along the direction of $$\vec v$$ and direction of $$\vec a_r$$ is radially inward or opposite of $$\vec R$$ and normal to $$\vec v$$. Rewriting above expression in terms of tangential and normal direction to any point on path of motion as, $$\vec a=R\frac{d\omega}{dt}\hat t-\omega \frac{dR}{dt}\hat n=a_t\hat t+a_n\hat n$$ Now, $$\quad \frac{dR}{dt}=\frac{dR}{d\theta}\frac{d\theta}{dt}=R\omega\$$ and $$\ \frac{d\omega}{dt}=\alpha$$

Thus, $$\quad\vec a=R\alpha\hat t-R\omega^2\hat n=a_t\hat t+a_n\hat n$$

Or, $$\quad a_t=R\alpha\$$ and $$\ a_n=- R\omega^2$$

• Why is $\frac{dR}{d\theta}=R$? Nov 5, 2022 at 16:57
• $dR$ is arc for circular motion, $dR=Rd\theta$. Nov 6, 2022 at 2:36

Short answer. Tangential acceleration makes the magnitude of the velocity change, while normal acceleration makes the direction of the velocity change.

(I really appreciated the approach of treating kinematics with vectors, and not to rely on a particular choice of a set of coordinates. IMHO it's the right approach)

Some details. The velocity of a point is tangent to its trajectory, and thus we can write the velocity vector as

$$\mathbf{v} = v \mathbf{\hat{t}}$$,

i.e. as the product of a unit vector $$\mathbf{\hat{t}}$$, that provides the information about the direction of the vector, and a scalar $$v$$, the magnitude of the velocity (the number you read on the tachometer of your car, as an example).

Now, we can take time derivative of the velocity, using the law for the derivative of a product (both the magnitude and the direction of the velocity depends on time, in general),

$$\dfrac{d \mathbf{v}}{dt} = \dfrac{d}{dt} (v \mathbf{\hat{t}}) = \underbrace{\dfrac{ dv}{dt} \mathbf{\hat{t}}}_{\mathbf{a}_t} + \underbrace{v \dfrac{d \mathbf{\hat{t}}}{dt}}_{\mathbf{a}_n}$$,

being the first term tangent to the trajectory (since it has the same direction as $$\mathbf{\hat{t}}$$), and the second term orthogonal to that, since it's in the direction of the time derivative of a unit vector, that is always orthogonal to the direction itself (or zero, if the direction doesn't change), as we can prove using

$$1 = |\mathbf{\hat{t}}|^2 = \mathbf{\hat{t}} \cdot \mathbf{\hat{t}}\qquad \rightarrow \qquad$$ $$0 = \dfrac{d}{dt}|\mathbf{\hat{t}}|^2 = 2 \mathbf{\hat{t}} \cdot \dfrac{d \mathbf{\hat{t}}}{dt}$$.

To get a deeper insight into normal acceleration, we need just a bit of differential geometry of curves in space. The time derivative of the unit tangent vector can be written as

$$\dfrac{d \mathbf{\hat{t}}}{dt} = k v \mathbf{\hat{n}} = \frac{v}{r} \mathbf{\hat{n}}\qquad \rightarrow \qquad\mathbf{a}_n = k v^2 \mathbf{\hat{n}} = \dfrac{v^2}{r} \mathbf{\hat{n}}$$

using the local unit normal vector $$\mathbf{\hat{n}}$$ (pointing towards the center of the local circle of curvature) and the local curvature of the trajectory $$k$$, or the local curvature radius $$r = \frac{1}{k}$$ (the radius of the local circle of curvature) so that we can recognize the more familiar centripetal acceleration in the normal acceleration. Eventually, we can write the acceleration as

$$\dfrac{d \mathbf{v}}{dt} = \mathbf{a}_t + \mathbf{a}_n = \dfrac{ dv}{dt} \mathbf{\hat{t}} + v \dfrac{d \mathbf{\hat{t}}}{dt} = \underbrace{\dfrac{ dv}{dt}}_{a_t} \mathbf{\hat{t}} + \underbrace{k v^2}_{a_n} \mathbf{\hat{n}}$$.

• Wonderful! Could you explain a little bit more what $k$ is please? Thanks very much! Nov 2, 2022 at 11:10
• Try to have a look at wiki, or here: basics.altervista.org/test/Math/differential_geometry/…. Are you familiar with the arc-length parametrization of curves, that makes things quite natural? Or, in other words, what's your proficiency in math and differential geometry, or what level are you attending? Nov 2, 2022 at 11:11
• I can definitely grasp that, I just have to look into it which I would do, but I thought this would make it easier for other people wondering the same thing who came upon this thread. But if it's too elaborate for this answer then that's ok :) I've done calc II but it's been a little while and I'm rehearsing some material now which is how I came to this question. Nov 2, 2022 at 11:28