## Quick Reference

Motion of a particle in a circular path. Suppose that the path of the particle *P* is a circle in the plane, with centre at the origin *O* and radius *r*_{0}. Let **i** and **j** be unit vectors in the directions of the positive *x*- and *y*-axes. Let **r**, **v** and **a** be the position vector, velocity and acceleration of *P*. If *P* has polar coordinates (*r*_{0}, *θ*), then

**r** = *r*_{0} (**i** cos *θ* + **j** sin *θ*),

**v** = **ṙ** = *r*_{0}(−*θ̇***i** sin *θ* + *θ̇***j** cos *θ*),

**a** = **r̈** = *r*_{0}(−*θ̈***i** sin *θ* − *θ̇*^{2}**i** cos *θ* + *θ̈***j** cos *θ* − *θ̇*^{2}**j** sin *θ*).

Let **e**_{r}=**i** cos *θ*+**j** sin *θ* and **e**_{θ}=−**i** sin *θ*+**j** cos *θ*, so that **e**_{r} is a unit vector along *OP* in the direction of increasing *r*, and **e**_{θ} is a unit vector perpendicular to this in the direction of increasing *θ*. Then the equations above become

r=*r*_{0}**e**_{r}, **v**=**ṙ**=*r*_{0}*θ̇***e**_{θ}, **a**=**r̈**=−*r*_{0}*θ̇*^{2}**e**_{r}+*r*_{0}*θ̈***e**_{θ}.

If the particle, of mass *m*, is acted on by a force **F**, where **F**=*F*_{1}**e**_{r}+*F*_{2}**e**_{θ}, then the equation of motion *m***r̈**=**F** gives−*mr*_{0}*θ̇*^{2}=*F*_{1} and *mr*_{0}*θ̈*=*F*_{2}. If the transverse component *F*_{2} of the force is zero, then *θ̇*=constant and the particle has constant speed.

See also angular velocity and angular acceleration.

*Subjects:*
Mathematics.