Derive the expressions for velocity and acceleration

+2 votes


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 by Ballysmo (44 points)
    

1 Answer

+1 vote

This explains really well what you are trying to do, hope it helps:

 

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.

 

Source: http://en.wikipedia.org/wiki/Centripetal_force

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

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