0
$\begingroup$

When I was studying uniform circular motion and thinking of ways to derive the expressions of related vectors like centripetal and centrifugal acceleration I noticed a certain oddity that occurred in my derivation. I want to know if it is simply a coincidence or whether there is some way to explain it.

Let us consider a circle of uniform radius $r$ with the $r$ making an angle of $\alpha$. Now let us consider two unit vectors $\vec e_r$ along the radial direction with the tail connected to the tip of the $r$ vector. Also the other is $\vec e_t$ where $\vec e_t$ is along the tangent of the circle.

Now we get on resolving them into the components;
$$\vec e_t = cos\alpha \hat j -sin\alpha \hat i$$ $$\vec e_r = cos \alpha \hat i + sin \alpha \hat j$$ $$ \vec r = \vert r \vert \cdot \vec e_r$$ Now differentiating w.r.t time, we can derive all off the expressions required however I noticed that the $\vec e_r$ when differentiated w.r.t angle $\alpha$ we get,
$$ \frac {d e_r}{d \alpha} = \vec e_t$$ Is this simply a coincidence or is there some sort of concept that can be unearthed here?

Also, can someone tell me someway I can create an image to make the figure clear?

$\endgroup$
0
$\begingroup$

The tangent vector $\vec{T}(\alpha)$ for a curve parameterized by some parameter $\alpha$, say $\vec{r}(\alpha)$, is given by the derivative with respect to $\alpha$, $$\vec{T}(\alpha)\equiv\frac{d}{d\alpha}\vec{r}(\alpha).$$ Your radial unit vector $\hat{e}_r(\alpha)$ is just the unit length vector pointing to the point on the circle parameterized by $\alpha$, so it's natural that taking the $\alpha$ derivative of $\hat{e}_r$ gives you $\hat{e}_t$, the unit vector along the tangent to the circle.

$\endgroup$
  • $\begingroup$ Why is the tangent to the curve parameterized by some parameter $\alpha$ given by $\frac {d}{d \alpha}$ $\vec r (\alpha)$? Or is that a question for a mathematics forum? $\endgroup$ – Prakhar Nagpal Jul 1 '18 at 15:48
0
$\begingroup$

It seems that this formula just reflects, in the vector form, a relationship between the length of an arc, the radius and the angle subtended by the arc: $S=r\theta$.

The diagram below, shows two unit radial vectors, $\vec e_r$ and $\vec e_r+d\vec e_r$, separated by a small angle $d\alpha$. It also shows a difference vector, $d\vec e_r$, pointing roughly in the direction of a unit tangent vector $\vec e_t$.

enter image description here

With respect to this diagram, your formula says that the magnitude of the difference vector, $d\vec e_r$, is equal to the product of the unit radius, $1$, and the angle $d\alpha$ ($S=r\theta$), while its direction coincides with the direction of the tangent vector $\vec e_t$, which is a good approximation for small angles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.