Waste less time on Facebook — follow Brilliant.

Deriving Formula for Centripetal Force

The conditions for a centripetal force to be present is circular motion, specifically when the angular frequency and speed are constant. First, start off with the vector or parametric equation for circular motion about the origin. The radius of the circular path is r meters, the angular frequency of the particle is omega.

\( \mathbf{s}=\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}r \cos\left ( \omega t \right )\\ r \sin\left ( \omega t \right ) \end{pmatrix}\)

The definition of velocity is the rate of change in displacement, so to get velocity, differentiate the position vector for the particle with respect to time.

\(\mathbf{v}=\frac{d}{dt}\mathbf{s}=\begin{pmatrix}-r \omega \sin\left ( \omega t \right )\\ r \omega \cos\left ( \omega t\right )\end{pmatrix} \)

Velocity is not constant but if we take the magnitude, we find that its magnitude (speed) is, so the conditions for centripetal force to be present are satisfied.

\(|\mathbf{v}|=\sqrt{(r\omega)^2 \cos^2 \left ( \omega t \right ) +(r\omega)^2 \sin^2 \left ( \omega t \right )}\)

\(|\mathbf{v}|=r\omega\sqrt{\cos^2 \left ( \omega t \right ) +\sin^2 \left ( \omega t \right )}\)

\(\therefore |\mathbf{v}|=r\omega =r 2\pi f \)

Acceleration is the rate of change in velocity so differentiate the velocity vector to get the acceleration vector.

\(\mathbf{a}=\frac{d}{dt}\mathbf{v}=\begin{pmatrix}-r \omega^2 \cos\left ( \omega t \right )\\ -r \omega^2 \sin\left ( \omega t\right )\end{pmatrix} \)

With this expression for acceleration, we can show that acceleration is always perpendicular to velocity, another requirement for circular motion. Velocity is perpendicular to acceleration if and only if their scalar product is zero.

\(\mathbf{v}\perp\mathbf{a} \Leftrightarrow \mathbf{v}\cdot\mathbf{a}=0\)

\(\mathbf{v}\cdot\mathbf{a}=\begin{pmatrix}-r \omega \sin\left ( \omega t \right )\\ r \omega \cos\left ( \omega t\right )\end{pmatrix}\cdot\begin{pmatrix}-r \omega^2 \cos\left ( \omega t \right )\\ -r \omega^2 \sin\left ( \omega t\right )\end{pmatrix}=r^2\omega^3\cos \left ( \omega t \right ) \sin\left ( \omega t \right )-r^2\omega^3\cos \left ( \omega t \right ) \sin\left ( \omega t \right )=0\)

\(\therefore \mathbf{v}\perp\mathbf{a} \)

Moving on, the magnitude of acceleration is

\(|\mathbf{a}|=\sqrt{r^2 \omega^4 \sin^2 \left ( \omega t \right ) +r^2 \omega^4 \cos^2 \left ( \omega t \right )}=r \omega^2\)

Multiply by \( r\)

\(r|\mathbf{a}| =r^2 \omega^2=\left (r \omega \right ) ^2 \)

Substitute \( |\mathbf{v}|=r\omega \)

\(r|\mathbf{a}| =|\mathbf{v}|^2 \)

Divide by \( r\) and we have the formula for centripetal acceleration.

\(\therefore |\mathbf{a}| =\frac{|\mathbf{v}|^2}{r} \)

Assuming the particle has a constant mass, multiply both sides by \(m\) and substitute the left hand side for the magnitude of the force vector because of Newton's second law. Thus we have the formula for centripetal force.

\(\therefore |\mathbf{F}| =m\frac{|\mathbf{v}|^2}{r} \)

Note by Jack Han
1 year, 3 months ago

No vote yet
1 vote


There are no comments in this discussion.


Problem Loading...

Note Loading...

Set Loading...