Centripetal force

Centripetal acceleration is a force that acts on a body that revolves around a centre of a circular path. This force keeps the body in circular motion. Newton's first law is that a moving object will continue in its motion in a straight path unless acted on by an external force.

Athlete must pull to make his hammer throw work properly

When you hold a rope with an object attached to it, and spin it in a circular motion around the hand of the person holding the rope, the rope becomes tight and keeps the body from flying away. This force acts towards the center of motion, not away from it.

An object changes direction, but may maintain a constant speed, if acted upon by a centripetal acceleration. In such a case, we know from its angular frequency that ${\textstyle v=\omega r}$. From Newton's second law we know that force equals mass times acceleration. Here acceleration is the negative of the radius coming from the centre of the circular motion. $F_{r}=ma_{r}=-mv^{2}{\frac {\vec {\mathbf {r} }}{r^{2}}}=-{\frac {mv^{2}}{r}}{\hat {\mathbf {r} }}$

In this formula, think that the centripetal force and the centripetal momentum are the same thing because they are both measured in Newtons.

If, however, the angular frequency is not constant (as might be the case in a roller-coaster ride) then ${\textstyle \omega }$ does vary with respect to time. Here

${\frac {d\omega }{dt}}={\frac {d^{2}\theta }{d\theta ^{2}}}={\frac {1}{r}}{\frac {dv}{dt}}$

Thus

$r={\frac {s}{\theta }}={\frac {dv}{d\omega }}$

as ${\textstyle \omega ={\frac {d\theta }{dt}}}$ and ${\textstyle s=r\theta }$ , which we know from the angular frequency.

Geometric proof for uniform circular motion

Uses vector subtraction and similar triangles to get a=v^2/R

In the figure to the right we define the displacement vector ${\vec {s}}$ to represent motion in a circle. The magnitude of ${\vec {s}}$ is denoted as $R\equiv ||{\vec {s}}||$ and represents the radius of the particle's orbit.