Having trouble with derivation in Non-Uniform Circular, need help! (1 Viewer)

emilios · Apr 16, 2014

I have the correct expressions for x(double dot) and y(double dot) after using the product rule. I need the correct expressions for the tangential and normal accelerations.

The textbooks I'm using are assuming I already know how to resolve forces.... but I don't.

If someone could explain it to me like Im 5 that would be greatly appreciated, I'm really struggling with non-uniform stuff.

Sy123 · Apr 16, 2014

I believe non-uniform circular motion is not the HSC syllabus, however:

$\omega = \frac{d\theta}{dt} \ , \ \phi = \frac{d\omega}{dt}$

$x = r\cos \theta \ \Rightarrow \ \dot{x} = - r\omega \sin \theta \ \Rightarrow \ \ddot{x} = -r\phi \sin \theta - r\omega^2 \cos \theta$

$y = r\sin \theta \ \Rightarrow \ \dot{y} = r\omega \cos \theta \ \Rightarrow \ \ddot{y} = r\phi \cos \theta - r\omega^2 \sin \theta$

$\\ a = \sqrt{\ddot{x}^2 + \ddot{y}^2} = r\sqrt{\phi^2 \sin^2 \theta + 2\phi \omega^2 \sin \theta \cos \theta + w^4 \cos^2 \theta + \phi^2 \cos^2 \theta - 2\phi \omega^2 \sin \theta \cos \theta + w^4 \sin^2 \theta} \\ \\ = r\sqrt{\phi^2 + \omega^4}$

Notice that if $\phi = 0$ (angular acceleration being 0, i.e. uniform circular motion, we get the familiar formula a=rw^2)

emilios · Apr 16, 2014

Oh whoops I already arrived at that expression. I thought it had a more elegant form. Thanks a lot for that mate!!!

braintic · Apr 17, 2014

Sy123 said:
I believe non-uniform circular motion is not the HSC syllabus

It is.

(dot point from syllabus) prove that the tangential and normal components of the force acting on a particle moving in a circle of radius R with angular velocity ω need to be mRω(dot) and –mRω^2 respectively.

The non-zero tangential component signifies non-uniform motion.

There have been HSC questions involving motion in a vertical circle (which is non-uniform).

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Having trouble with derivation in Non-Uniform Circular, need help! (1 Viewer)

emilios

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Sy123

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emilios

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braintic

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