Circular Motion


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

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 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 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


This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.


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