Circular motion

Your browser does not support the canvas element. $$\large \vec{r}=R\cos(\omega t)\,\hat{x} + R\sin(\omega t)\,\hat{y}.$$ $$\large \vec{v}=-\omega R\sin(\omega t)\,\hat{x} + \omega R\cos(\omega t)\,\hat{y}.$$ $$\large \vec{a}=-\omega^2 R\cos(\omega t)\,\hat{x} - \omega^2 R\sin(\omega t)\,\hat{y} =-\omega^2\vec{r}.$$ $$\large \vec{F}=-m\omega^2 R\cos(\omega t)\,\hat{x} - m\omega^2 R\sin(\omega t)\,\hat{y}=-m\omega^2\vec{r}= -m\frac{|\vec{v}|^2}{R^2}\vec{r}.$$

The position vector for a particle moving in a circle is,

$$\vec{r}=R\cos(\omega t)\,\hat{x} + R\sin(\omega t)\,\hat{y}.$$

The circular motion (represented by the red ball) is a superposition of a harmonic motion $R\cos (\omega t)$ in the $x$-direction (represented by the blue ball) and a harmonic motion $R\sin (\omega t)$ in the $y$-direction (represented by the green ball).

Using the identity $\sin^2\theta+\cos^2\theta =1$, the length of the position vector can be shown to be independent of time $|\vec{r}|=R$,

$$|\vec{r}|=\sqrt{R^2\cos^2(\omega t)+R^2\sin^2(\omega t)}=R.$$

The velocity vector is the derivative of the position vector,

$$\vec{v}=-\omega R\sin(\omega t)\,\hat{x} + \omega R\cos(\omega t)\,\hat{y}.$$

The length of the velocity vector is, $|\vec{v}|=\omega R$. Since $\omega = \frac{2\pi}{T}$, where $T$ is the period, the velocity is the circumference of the circle divided by the time it takes to go once around the circle, $|\vec{v}|=\frac{2\pi R}{T}.$

The position vector points radially outwards from the center of the circle and the velocity vector is tangential to the circle. The position vector and the velocity vector are perpendicular to each other, $\vec{r}\cdot\vec{v}=-\omega R^2\cos(\omega t)\sin(\omega t)+ \omega R^2\cos(\omega t)\sin(\omega t) = 0$.

The acceleration vector is the derivative of the velocity vector,

$$\vec{a}=-\omega^2 R\cos(\omega t)\,\hat{x} - \omega^2 R\sin(\omega t)\,\hat{y}.$$

The acceleration vector is proportional to the position vector $\vec{a} = -\omega^2\vec{r}$ and points radially inwards. The acceleration vector is also perpendicular to the velocity $\vec{a}\cdot\vec{v}=0$, and has a length, $|\vec{a}|= \omega^2R$.

The force is the mass times the acceleration, $\vec{F}=m\vec{a}$, and is proportional to the position vector. It points radially inwards.

$$\vec{F}=-m\omega^2 R\cos(\omega t)\,\hat{x} - m\omega^2 R\sin(\omega t)\,\hat{y}=-m\omega^2\vec{r}.$$

The length of the force vector is $|\vec{F}|=m\omega^2R=\frac{mv^2}{R}$. The force needed to keep a particle moving in a circle is called the centripedal force. This force increases with the square of the angular frequency $\omega$.