Radial acceleration in uniform circular motion

[andyrk]

Why is there only a radial component of acceleration present if a body is undergoing uniform circular motion whereas in non uniform circular motion both tangential and radial component of acceleration are present?

 

[tia89]

The reason is the following: in circular motion there will always be some acceleration,because velocity always changes; even in uniform motion (where velocity does not change in modulus) it will at least change direction. And thus an acceleration is present, as acceleration is defined as change of velocity in time.
Now, clearly velocity is always tangential to the motion (otherwise it would not be circular) and therefore if there is also a component of acceleration along the direction of velocity, then also the modulus of velocity will change, and we have non uniform motion, while in case the acceleration is purely orthogonal to velocity, there will be no component able to change the modulus of velocity, and only the direction will change; this is uniform motion.

 

[andyrk]

Uhmm..I think I still didn't get it. How can we say that if the magnitude(modulus as you say) is not being changed then there will be no tangential acceleration? It is similar to saying that
(1)Only a radial component of acceleration can change the direction of velocity in circular motion.
(2)Only a tangential component of acceleration can change the magnitude(modulus) of velocity in circular motion.

How are both of the above statements true? I mean is there any proof for them? Prove the above statements?

 

[voko]

The square of the magnitude of velocity is related to kinetic energy. Any force doing work on a particle changes its kinetic energy. A radial force does no work, because it is perpendicular to motion. So it does not change the kinetic energy, so it does not change the magnitude of velocity. Ergo, the only way to change the magnitude is via a tangential force.

 

[andyrk]

Thanks!

 

[andyrk]

Why can it be only a tangential force? I mean any force other than radially inward or radially outward can change the velocity of the particle then. Then why does it specifically have to be a tangential force which corresponds to a tangential acceleration?

 

[dreamLord]

It can be any force which has a component along the tangential direction. Any force which is perpendicular to the (instantaneous) velocity will not change it's magnitude. All other forces will have a component along the tangential direction.

 

[andyrk]

It can be any force which has a component along the tangential direction. Any force which is perpendicular to the (instantaneous) velocity will not change it's magnitude. All other forces will have a component along the tangential direction.

Then why does it have to be a tangential force? Because in non uniform circular motion velocity changes and there are one radial and one tangential component of acceleration. Why does the latter one have to be tangential?

 

[tia89]

in non uniform circular motion velocity changes and there are one radial and one tangential component of acceleration. Why does the latter one have to be tangential?

Just because when you have a general acceleration, you decide (for convenience) to split it in a component perpendicular to the motion and one tangential...
You could do otherwise, but nothing changes, apart from the fact that doing different would be much more unusual and not so helpful

 

[voko]

Then why does it have to be a tangential force? Because in non uniform circular motion velocity changes and there are one radial and one tangential component of acceleration. Why does the latter one have to be tangential?

Any force can be resolved into two components: tangential and normal. The tangential component is parallel to velocity, the normal is perpendicular. "Radial" in circular motion is a particular example of "normal". As explained above, the normal component cannot change the magnitude of velocity. So it can only be done by the tangential component. Conversely, the tangential component, obviously, cannot change the direction of velocity, so it can only be done by the normal component.

 

[dreamLord]

In non uniform circular motion the radial component does not change the magnitude of the velocity. Voko explained this in post #4. Think of it as this way : you have a car which has stopped working. Your friend is in the driver's seat and is controlling the steering wheel (hence taking care of the radial acceleration required for circular motion). You are pushing the car from behind. The road along which you have to travel is exactly circular. To make the car go faster, how will you push the car? You will push it in the direction in which the car is instantaneously moving - which is tangential.

I don't know if this example is 'real life' enough. Apologies.

 

[tia89]

What you do usually is the following... for a generic acceleration, you split it in orthogonal components, one directed along the motion and one perpendicular and directed towards the center of the trajectory (which being circular does have a center... anyway you can do the same for a generic trajectory, just approximating locally the trajectory with the circle tangent to the trajectory). Then you see that the orthogonal (radial) component acceleration is the one responsible for the shape of the motion, while the tangential component (parallel to velocity then) is the one responsible for change in magnitude of the velocity.
You could also split in a different way (you can always decompose a vector in components according to some basis) but this above is the only convenient one to study the motion, other decompositions would be just theory

 

[Chestermiller]

Why don't you use the math to help explain it for you? Use polar coordinates. Let the velocity be $$\vec{v}=V(t)\vec{i_θ}$$
where V(t) is the magnitude of the tangential velocity and $\vec{i_θ}$ is the unit vector in the tangential direction. Take the derivative of the velocity with respect to time to get the acceleration vector, noting that
$$\frac{d\vec{i_θ}}{dθ}=-\vec{i_r}$$ and $$\frac{dθ}{dt}=V/R$$ where $-\vec{i_r}$ is the unit vector in the radial direction and R is the radius of the circle.