# HELP proof that a=v^2/r in NON-uniform circular motion

1. Nov 5, 2011

### iampaul

HELP!!!proof that a=v^2/r in NON-uniform circular motion

Most of the proofs I have read especially the geometric approach assume that the speed is constant, so it mustn't apply for non-uniform circular motion. Some proofs use calculus and starts from the position vector and differentiates it to get the velocity and differentiates again to get the acceleration vector. What i can't understand with this proof is that the magnitude of the acceleration vector becomes exactly equal to v2/r even if it wasn't assumed in the proof that the speed is constant. If the speed isn't constant then the magnitude of the acceleration vector must be more than that. Here is the link for the proof: http://www.youtube.com/watch?v=YRBRarbMCyE&feature=relmfu
Is this proof just for uniform circular motion??

2. Nov 5, 2011

### superg33k

Re: HELP!!!proof that a=v^2/r in NON-uniform circular motion

This calculation is only for a object moving in circular motion.

If an object is moving at a constant speed then it goes in a straight line. If however it keeps changing direction then it must be accelerating in the new direction. An object moving in a circular direction, as in the video you posted, must always be accelerating. In fact objects in circular motion are always accelerating towards the centre of rotation.

Think of the Earth orbiting the Sun. The force of gravity is causing us to accelerate towards the Sun, but our initial velocity is causing us to orbit it instead of crashing into it. The video is showing the calculation for the central acceleration.

3. Nov 5, 2011

### AlephZero

Re: HELP!!!proof that a=v^2/r in NON-uniform circular motion

The radial component of the acceleration is v2/r for both uniform and non-uniform circular motion.

For non-uniform motion there is also a tangential component of the acceleration, so the direction of the acceleration vector is not towards the center of the circle and its magnitude is not v2/r.

The radial component of the acceleration comes from the change in direction. The tangential component comes form the change in speed and its magnitude is just dv/dt.

4. Nov 6, 2011

### iampaul

Re: HELP!!!proof that a=v^2/r in NON-uniform circular motion

What I can't understand is how the magnitude of the acceleration vector became exactly equal with v^2/r, if the calculation in the video is for both uniform and non-uniform circular motion. The calculation seems to be only for uniform circular motion because the computed magnitude of the overall acceleration = v^2/r which is supposed to be the centripetal acceleration only so that there is no tangential acceleration. If it also applies for non-uniform circular motion as well, then shouldn't the overall acceleration be more than v^2/r because the magnitude of the overall acceleration is equal to sqrt(ac^2+at^2) and ac=V^2/r ? I'm really confused! Do you know where i can find a formal proof for
non-uniform circular motion? ThanKs!!!

5. Nov 6, 2011

### iampaul

Re: HELP!!!proof that a=v^2/r in NON-uniform circular motion

If it only applies for uniform circular motion what i can't understand is ..on what part of his proof did he assume that the speed is constant.. he only said in 3:37 that the speed isn't changing but it didn't appear in his proof...can someone please explain it to me... thanks again!!

6. Nov 29, 2011

### cmartin256

Re: HELP!!!proof that a=v^2/r in NON-uniform circular motion

Let $\vec p$ be the position vector of the particle undergoing circular motion, and $\theta$ be the angle subtended by the position vector and the horizontal.

$\vec p = r<cos \theta, sin \theta>$
$\vec v = \frac{d \vec p}{d t} = \frac{d \theta}{d t} \frac{d \vec p}{d \theta} = r \frac{ d \theta}{d t} <-sin \theta, cos \theta>$
$\vec a = \frac{d \vec v}{d t} = \frac{d \theta}{d t} \frac{d \vec v}{d \theta} = -r {\frac{d \theta}{d t}}^2 <cos \theta, sin \theta> + r \frac{d^2 \theta}{{d t}^2}<-sin \theta, cos \theta>$

$\vec {a_c}$ - centripital acceleration

$\vec{a_c} = -r {\frac{d \theta}{d t}}^2 <cos \theta, sin \theta>$ (the other part acts in the same direction as velocity, therefore tangentially. This part acts in the same opposite direction to position vector, hence centripetally)

$v = \frac{d \theta}{d t} r$

$a_c = |\vec{a_c}| = |-r \frac{v^2}{r^2} <cos \theta, sin \theta>|$
$a_c = r \frac{v^2}{r^2} = \frac{v^2}{r}$

7. Nov 29, 2011

### sophiecentaur

Re: HELP!!!proof that a=v^2/r in NON-uniform circular motion

The video is only dealing with the centripetal acceleration, which is just dependent upon the instantaneous velocity and the radius of curvature at that time. Any change in angular velocity will only result in a tangential acceleration - because 'he' showed that the only acceleration at right angles to the path is rωsquared. Remember that there was nothing in the derivation to show that angular acceleration actually contributes to the centripetal acceleration. If you keep the value of r constant and vary ω the only change in velocity, due to the change in ω can be tangential because <v> = <r> X <ω> and dr/dt is zero. This implies that the only change in v is at right angles to the axis of rotation and the position vector. (i.e. tangential) I think that's rigorous enough.
The overall acceleration will be the vector sum of the two accelerations, of course. So this describes not only non-uniform motion in a circle but any curvilinear (posh word) motion - such as an elliptical orbit round a planet - in which the only force will be gravitational - but this force can speed up or slow down the satellite (tangential acceleration) and give it a centripetal acceleration which is at right angles to its instantaneous direction of travel and depends on the instantaneous radius of curvature. The radius will be changing all the time in this case.