Vector form of centrepetal 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}

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.