Engineer Physics/Force/Centripetal

$F=m{\frac {v^{2}}{r}}$

Formulation

The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:

$$F_{c}=ma_{c}={\frac {mv^{2}}{r}}$$

$$a_{c}=\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta t}}$$

where $a_{c}$ is the centripetal acceleration and $\Delta {\textbf {v}}$ is the difference between the velocity vectors. Since the velocity vectors in the above diagram have constant magnitude and since each one is perpendicular to its respective position vector, simple vector subtraction implies two similar isosceles triangles with congruent angles – one comprising a base of $\Delta {\textbf {v}}$ and a leg length of $v$, and the other a base of $\Delta {\textbf {r}}$ (position vector difference) and a leg length of $r$

$${\frac {|\Delta {\textbf {v}}|}{v}}={\frac {|\Delta {\textbf {r}}|}{r}}$$

$$|\Delta {\textbf {v}}|={\frac {v}{r}}|\Delta {\textbf {r}}|$$

Therefore, $|\Delta {\textbf {v}}|$ can be substituted with ${\frac {v}{r}}|\Delta {\textbf {r}}|$

$$a_{c}=\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {v}}|}{\Delta t}}={\frac {v}{r}}\lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta t}}=\omega \lim _{\Delta t\to 0}{\frac {|\Delta {\textbf {r}}|}{\Delta t}}=v\omega ={\frac {v^{2}}{r}}$$

The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle (the circle that best fits the local path of the object, if the path is not circular). The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle, related to the tangential velocity by the formula

$$v=\omega r$$

so that

$$F_{c}=mr\omega ^{2}\,.$$

Expressed using the orbital period T for one revolution of the circle,

$$\omega ={\frac {2\pi }{T}}$$

the equation becomes

$$F_{c}=mr\left({\frac {2\pi }{T}}\right)^{2}.$$

In particle accelerators, velocity can be very high (close to the speed of light in vacuum) so the same rest mass now exerts greater inertia (relativistic mass) thereby requiring greater force for the same centripetal acceleration, so the equation becomes

$$F_{c}={\frac {\gamma mv^{2}}{r}}$$

where

$$\gamma ={\frac {1}{\sqrt {1-{\frac {v^{2}}{c^{2}}}}}}$$

is the Lorentz factor.

Thus the centripetal force is given by:

$$F_{c}=\gamma mv\omega$$

which is the rate of change of relativistic momentum $\gamma mv$.