# Given the mass and the velocity of an object in circular motion with radius r, how do we calculate the magnitude of the centripetal force, F?

Mar 3, 2016

Constant acceleration (change of direction) requires a constant force, which is described by $F = \frac{m {v}^{2}}{r}$ if $v$ is measured in $m {s}^{-} 1$ or $F = m {\omega}^{2} r$ if $\omega$ is measured in $r a {\mathrm{ds}}^{-} 1$.

#### Explanation:

Acceleration is defined as the rate of change of velocity, and velocity has a direction.

Since the direction of something in circular motion is always changing, it requires a constant acceleration, and Newton's Second Law indicates that constant acceleration requires a constant force.

$F = m a$

If we look at the instantaneous linear velocity of the object at any moment as $v$ $m {s}^{-} 1$, the expression for the centripetal acceleration is:

$a = {v}^{2} / r$

So the force acting is simply $F = \frac{m {v}^{2}}{r}$

Let's look at where that expression for the acceleration comes from.

The speed of the object is constant: it is not accelerating because its speed increases (speeding up) or decreases (slowing down). It is accelerating because its direction is changing constantly.

If we imagine a direction vector, which is a tangent to the circle in which the object is moving, at each moment it points in a slightly different direction.

Acceleration is just change in velocity divided by change in time:

$a = \frac{\Delta v}{\Delta t}$

This Khan Academy video offers a more-detailed explanation of how we get from there to $a = {v}^{2} / r$: https://www.khanacademy.org/science/physics/centripetal-force-and-gravitation/centripetal-acceleration-tutoria/v/visual-understanding-of-centripetal-acceleration-formula

If, instead, we measure the radial velocity of the object in radians per second $\left(r a {\mathrm{ds}}^{-} 1\right)$, the expression is:

$F = m {\omega}^{2} r$