Vector form of centripetal force

Question

We know the centripetal force $F_c$ had magnitude $m\omega^2r$. But let's try to write it in vector form.

First of all,since it is directed along the radius,the unit vector in radial direction in this case is $-\hat{\boldsymbol{r}}$. And since the magnitude is $m\omega^2r$,we finally get $$\vec{\mathrm{F_c}}=-\mathrm{m}\mathrm{\omega}^2\mathrm{r}\hat{\boldsymbol{r}}=-\mathrm{m}\frac{v^2}{r^2}\vec{\mathrm{r}}.$$

But this is not how it us done in the books. According to them,$$\vec{\mathrm{F_c}}=-\mathrm{m}\frac{v^2}{r^3}\vec{\mathrm{r}}.$$

I don't understand how they got it,the one i did seems to be completely fine to me. Could anyone tell me the mistake i made?

Answer 1

In times of doubt it is nice to have a simple example on which we can rely. Start with uniform circular motion: \begin{align} \vec r&=r(\cos(\omega t),\sin(\omega t))\\ \vec v&=r\omega(-\sin(\omega t),\cos(\omega t))\\ \vec a&=-r\omega^2(\cos(\omega t),\sin(\omega t)) \end{align} From the second formula we can derive $v=r\omega\implies\omega=\frac{v}r$. We can write the last formula as $\vec a=-\omega^2\vec r$. Combining these two gives \begin{align} \vec a&=-\frac{v^2}{r^2}\vec r\\ \vec F_C&=-\frac{mv^2}{r^2}\vec r \end{align}

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Edit in response to Nico: The units of $\vec r$ are distance, e.g. meters. If you want to reproduce wikipedia's formula more closely, you can decompose $\vec r$ into $$\vec r=r\hat r,$$ where $r=|\vec r|$ is the length of $\vec r$ and $\hat r$ is a unit vector defined by $\hat r=\tfrac 1 r\vec r$. The length of the unit vector is 1 and it has no units. Using this decomposition, you can write the centripetal force as $$\vec F_C=-\frac{mv^2}{r}\hat r$$

Take away from this: if you want to write down the vectorized version of some scalar equation, you can more closely reproduce the scalar equation by using unit vectors. When you want to actually compute stuff, the form that uses regular vectors can be easier to work with.

Answer 2

The formula given in the book is obviously wrong, and the dimensions it-self are incorrect. You are not wrong, the book is.

Answer 3

The vector form is $$ \vec{F} = m \left( \vec{\omega} \times (\vec{\omega} \times \vec{r}) \right) \tag{1}$$

where $m$ is the mass of the object, $\vec{r}$ is the position of the center of mass relative to the axis of rotation, and $\vec{\omega}$ is rotational (orbital) velocity of the body.

This is derived from the velocity of the center of mass when rotating about a fixed axis

$$ \vec{v} = \vec{\omega} \times \vec{r} \tag{2}$$

where $\times$ is the vector cross product. In a rotating frame, the cross product provides the changes in a vector due to the rotation of the local directions.

Then the change in velocity (acceleration vector) which is derived from the time derivative of velocity is

$$ \vec{a} = \vec{\omega} \times \vec{v} = \vec{\omega} \times ( \vec{\omega} \times \vec{r}) \tag{3}$$

Finally, the force required to provide this acceleration is $\vec{F} = m \vec{a}$ and hence (1).

Answer 4

Here, $\hat{\mathbf{r}}$ is a unit vector in radial direction whereas, $\vec{\mathbf{r}}$ is a position vector.

Also, $\vec{\mathbf{r}} = |\vec{\mathbf{r}}|\hat{\mathbf{r}}$.

This implies, $\vec{F} = -\frac{mv^2}{r^3}\vec{\mathbf{r}} = -\frac{mv^2}{r^3}|\vec{\mathbf{r}}|\hat{\mathbf{r}} = -\frac{mv^2}{r^2}\hat{\mathbf{r}} $.

The mistake you did is change $\hat{\mathbf{r}}$ to $\vec{\mathbf{r}}$ in your first equation.