tangential acceleration for uniform circular motion

Question

I understand that circular motion is defined by 2 components of acceleration, one tangential and one radial and their resultant is what causes circular motion.

I am confused though as to why it is said that in uniform circular motion (constant angular velocity) there is no tangential acceleration. How can the radial vector be always pointing towards the centre? Is there is no tangential vector component to it?

So is there still a vectorial tangential acceleration (since the vector points to a different point at each time)?

And is centripetal acceleration the resultant of the tangential and radial accelerations?

Answer 1

Radial acceleration $\vec a_{rad}$ takes care of turning (when pulling perpendicular to the velocity vector $\vec v$, it can only turn it, not increase it), and tangential acceleration $\vec a_{tan}$ takes care of speeding up (when pulling parallel to $\vec v$, it can only increase it, not turn it).

1. A car speeding up while driving straight, has a $\vec a_{tan}$ but no $\vec a_{rad}$.
2. A car turning but not speeding up has a $\vec a_{rad}$ only and no $\vec a_{tan}$.
3. A car speeding up while turning has both a $\vec a_{rad}$ and a $\vec a_{tan}$.

A uniform circular motion is case 2. The word "uniform" means constant speed.

Answer 2

For uniform circular motion, the position vector in time $t$ is given by: $$\vec{r}=R\sin(\omega t) \vec{e}_x+R\cos(\omega t) \vec{e}_y,$$ where $R$ is the radius, $\omega$ the (constant) angular velocity and $\vec{e}_x$, $\vec{e}_y$ the orthonormal unit vectors for $x$ and $y$.

The acceleration vector is given by: $$\vec{a}=\frac{\mathrm{d^2}\vec{r}}{\mathrm{d}t^2}$$ Carry out the two derivations and find: $$\vec{a}=-\omega^2[R\sin(\omega t) \vec{e}_x+R\cos(\omega t) \vec{e}_y]$$ $$\implies \vec{a}=-\omega^2 \vec{r}$$ So in uniform circular motion the only acceleration vector always points opposite to the position vector, to the centre of the trajectory.

If $\omega=f(t)$ (non-uniform circular motion) then the drivation changes, of course.

Answer 3

The radial component of the acceleration is the one pointing to the center. If the acceleration points to the center, then there is no tangential component. So I'm not really sure what you're asking.

Think about it again, and take care to distinguish the acceleration and the velocity vectors. The acceleration points to the center which makes the velocity point in tangential direction. Maybe that's where your confusion is coming from?

Centripetal acceleration is just another word for acceleration in direction of the center.

Answer 4

Elliptical orbits have tangential components so the speed changes.

Answer 5

Note a conker rotating on a string has a tension force pulling it toward the centre. That force tells us it also has acceleration directed toward centre! Dont confuse acceleration with velocity. to make something travel in a curve you need to shove or deflect the velocity vector slightly and repeat again and again to get a uniform circular motion. The little shoves are always sideways,, and sideways to the tangents points radially.

There are 2 ways (or more) of thinking about the components of acceleration. The Cartesian components of uniform circular motion are out of phase and oscillate result in giving the motion in a circle. However if we resolve into a instantaneous radial and tangential components then for uniform motion we only need a central or radial acceleration and the tangential is zero (not speeding up rotationally).