A particle is moving in a circle of radius r with constant angular velocity `omega`. At any point (r,`theta`) on its path , its position vector is `vec(r )` = r cos`theta hat(i) + r sin theta hat(j)`. Show that the velocity of the particle has no component along the radius.

Explanation

`vec(r ) = rcos theta hat(i) + r sin theta hat(j)`
`therefore" "(d hat(r ))/(dt) = -r sin theta (d theta)/(dt)hat(i) + r cos theta (d theta)/(dt) hat(j)`
= ` r omega(- sin theta hat(j) + cos theta hat(j))`
or, v = `|(d hat(r ))/(dt)| = sqrt(r^(2) omega^(2)) = r omega`
So, `vec(r )cdot(d hat(r ))/(dt) = r^(2) omega( - cos theta sin theta + cos theta sin theta ) = `0
As, `vec(r ) and (d hat(r ))/(dt)` are mutually perpendicular, the
component of `(d hat( r))/(dt)` along `vec(r )` is zero.