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# Circular Motion

#### Contents

##### Position

Consider a point on the unit circle whose position is given by the radius vector {$\mathbf r .$} The angle {$\mathbf r$} forms with the x-axis may be called {$\theta .$}

{$\mathbf r$} may be expressed as its x and y components {$\mathbf r = (r_x,r_y) = (\cos \theta, \sin \theta) .$}

That this is on the unit circle can be verified because the length of {$\mathbf r$} is 1, and its tail is on the origin. {$$\lVert \mathbf r \rVert = \sqrt{r_x^2 + r_y^2} = \sqrt{\cos^2\theta + \sin^2\theta} = \sqrt{1} = 1$$}

Acceleration not to scale
##### Motion

Now, the position of a particle in circular motion is a function of time, and clearly the angle of the vector with the x-axis is a function of time. (By convention, counter-clockwise rotation is taken as positive.)

{$\large \mathbf r(t) = (\cos \theta(t), \sin \theta(t) )$}

{$\omega$} is called the angular frequency. Its units are radians per second. This is the unit physics more commonly uses to express frequency than Hertz (Hz). Hz is cycles/second, so {$\omega = 2\pi f.$} For {$\theta$} expressed in radians (as it always is in physics) {$\theta = \omega t .$}

{$\large \mathbf r(t) = (\cos \omega t , \sin \omega t )$}
{$\large r_x(t) = \cos \omega t$}
{$\large r_y(t) = \sin \omega t$}

###### Velocity

Velocity is the instantaneous change in position, so it is the derivative with respect to time of the position.

{$$\large \dot r_x = \frac{d(\cos \omega t)}{dt} = -\omega\sin\omega t \\ \large \dot r_y = \frac{d(\sin \omega t)}{dt} = \omega\cos\omega t \\ \large \mathbf{v} = \mathbf{\dot r} = (-\omega\sin\omega t,\omega\cos\omega t)$$}

It is intuitively obvious that the velocity is perpendicular to the radius vector, but this can be verified by taking the dot product.

{\large \begin{align} \mathbf{v} \cdot \mathbf{r} &= (-\omega\sin\omega t,\omega\cos\omega t) \cdot (\cos \omega t , \sin \omega t ) \cr &= -\omega\sin\omega t\cos \omega t + \omega\cos\omega t\sin \omega t \cr &= 0 \ \tag*{ \blacksquare} \end{align}}

The magnitude of the velocity:

{\begin{align} \lVert \mathbf v \rVert &= \sqrt{v_x^2 + v_y^2} \\ &= \sqrt{(-\omega\sin\omega t)^2 + (\omega\cos\omega t)^2} \\ &= \sqrt{\omega^2\sin^2\omega t + \omega^2\cos^2\omega t}\\ &= \sqrt{\omega^2(\sin^2\omega t + \cos^2\omega t)} \\ &= \sqrt{\omega^2(1)} \\ &= \omega \end{align}}

###### Acceleration

Acceleration is the instantaneous change in velocity, so it is the derivative with respect to time of the velocity and the second derivative of position.

{$$\large a_x = \dot v_x = \ddot r_x = \frac{d^2(\cos \omega t)}{dt^2} = \frac{d(-\omega\sin\omega t)}{dt} = -\omega^2\cos\omega t \\ a_y = \dot v_y = \ddot r_y = \frac{d^2(\sin \omega t)}{dt^2} = \frac{d(\omega\cos\omega t)}{dt} = -\omega^2\sin\omega t \\ \mathbf a = -\omega^2(\cos\omega t, \sin\omega t) = -\omega^2\mathbf r$$}

This last substitution saves the work of calculating the dot product and the magnitude. Since it has a negative sign, it is anti-parallel to the position vector and it has a magnitude of {$\omega^2$} because {$-\omega^2$} is just a scalar and we can observe its absolute value, which is just {$\omega^2 .$}

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Categories: Classical Mechanics

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