# Derive the expressions for velocity and acceleration

A particle P moves with a constant speed v around the circular path of radius r. Start with the expression r=rr1 for the position vector of P and derive the expressions for the velocity v and acceleration a of P. The radial and transvers unit vectors are r1 and theta1 respectively.

so I know that with cartesian coordinates the position equation would be
rcos(theta)i+rsin(theta)but after that I'm a little bit stuck. The teacher made some comment about recognizing that theta is dependent on time.

asked Sep 2, 2012 in Physics

+1 vote

#### Calculus derivation

In two dimensions the position vector $\textbf{r}$ which has magnitude (length) $r$ and directed at an angle $\theta$ above the x-axis can be expressed in Cartesian coordinates using the unit vectors $\hat{x}$ and $y-hat$:[10]

$\textbf{r} = r \cos(\theta) \hat{x} + r \sin(\theta) \hat{y}.$

Assume uniform circular motion, which requires three things.

1. The object moves only on a circle.
2. The radius of the circle $r$ does not change in time.
3. The object moves with constant angular velocity $\omega$ around the circle. Therefore $\theta = \omega t$ where $t$ is time.

Now find the velocity $\textbf{v}$ and acceleration $\textbf{a}$ of the motion by taking derivatives of position with respect to time.

$\textbf{r} = r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y}$
$\dot{\textbf{r}} = \textbf{v} = - r \omega \sin(\omega t) \hat{x} + r \omega \cos(\omega t) \hat{y}$
$\ddot{\textbf{r}} = \textbf{a} = - r \omega^2 \cos(\omega t) \hat{x} - r \omega^2 \sin(\omega t) \hat{y}$
$\textbf{a} = - \omega^2 (r \cos(\omega t) \hat{x} + r \sin(\omega t) \hat{y})$

Notice that the term in parenthesis is the original expression of $\textbf{r}$ in Cartesian coordinates. Consequently,

$\textbf{a} = - \omega^2 \textbf{r}.$

The negative shows that the acceleration is pointed towards the center of the circle (opposite the radius), hence it is called "centripetal" (i.e. "center-seeking"). While objects naturally follow a straight path (due to inertia), this centripetal acceleration describes the circular motion path caused by a centripetal force.

answered Sep 4, 2012 by ~Expert~ (1,479 points)