Instantaneous acceleration vector toward the concave side of a curved path

Question

Why does the instantaneous acceleration vector of a moving particle always point toward the concave side of a curved path? My college textbooks mention this as something obvious, without further discussion.

Answer 1

Because the non-tangential component of the acceleration always points toward the concave side. This is a mathematical result, with the proofs given, but to provide physical intuiton consider the non-tangential component of acceleration. This component doesn't affect the magntitude of the velocity vector, but changes its direction in a circular fashion. Hence, at an instant, you can approximate how the velocity vector changes by considering it to be in a circle of radius of curvature. Now, the non-tangential component of the acceleration vector obviously points towards the center of the circle, or to the concave side. Hence, the final acceleration vector is angled from the tangential towards the concave side. For a little of the math (fully shown in the link),

Now, since the velocity squared and the curvature functions are both positive, the non-tangential component of the acceleration is in the same direction as the normal vector, which by definition points towards the concave side of the function. Hence, the net acceleration function is angled from the tangent towards the concave side.

Answer 2

Cicero gave you the math; let me give you the picture.

Looking at this:

The blue blob is an object initially moving horizontally to the right. At point B I give it a push upwards; it will then follow a new trajectory.

You probably have no difficulty determining that the mass had to accelerate vertically to get its new direction.

Now make the steps smaller, and get lots of them. Instead of a single jump, the object will make many tiny jumps. In the limit, the jumps become a smooth curve. But still, the object gets greater vertical velocity - it is accelerating. And the direction of the acceleration is up - away from the convex side (towards the concave side) of the curve.

I hope the pictures help your intuition.

Answer 3

All the answers presented here have a strong influence of intuition that comes from physics, or they use the concept of curvature. And that makes them quite clear. But I think there is an answer that comes only from geometry and differential calculus that is distinguished by its simplicity and crystalline mathematical clarity. It is as follows:

Let $\phi(t)$ be the angle from the positive $x$-axis to $\mathbf{T}(t)$, and let $\mathbf{n}(t)$ be the unit vector that results when $\mathbf{T}(t)$ is rotated counter-clockwise through an angle of $\pi/2$. Since $\mathbf{T}(t)$ and $\mathbf{n}(t)$ are unit vectors, they can be expressed as $$ \mathbf{T}(t)=\cos\phi(t)\mathbf{i}+\sin\phi(t)\mathbf{j}, $$ and $$ \mathbf{n}(t)=\cos\left(\phi(t)+\pi/2\right)\mathbf{i}+\sin\left(\phi(t)+\pi/2\right)\mathbf{j}=-\sin\phi(t)\mathbf{i}+\cos\phi(t)\mathbf{j} $$ Note that in the intervals where $\phi(t)$ is increasing, the vector $\mathbf{n}(t)$ points toward the concave side of the curve (Fig. 1), and in the intervals where $\phi(t)$ is decreasing, it points in the opposite direction to the concave side (Fig 2.).

Now we derive $\mathbf T(t)$: $$ \frac{d\mathbf{T}}{dt}=\frac{d\mathbf{T}}{d\phi}\,\frac{d\phi}{dt}=(-\sin\phi\,\mathbf{i}+\cos\phi\,\mathbf{j})\frac{d\phi}{dt}=\mathbf{n}\frac{d\phi}{dt}. $$ But
$$ \text{$\frac{d\phi}{dt}> 0$ on intervals where $\phi(t)$ is increasing, and } $$ $$ \text{$\frac{d\phi}{dt} < 0$ on intervals where $\phi(t)$ is decreasing.} $$ Thus, $\frac{d\mathbf{T}}{dt}$ has the same direction as $\mathbf{n}(t)$ on intervals where $\phi(t)$ is increasing and the opposite direction on intervals where $\phi(t)$ is decreasing. Therefore, $\frac{d\mathbf{T}}{dt}$ points “inward” toward the concave side of the curve in all cases, and hence so does $\mathbf{N}(t)=\mathbf{T}'(t)/|\mathbf{T}'(t)|$. For this reason, $\mathbf{N}(t)$ is also called the inward unit normal when applied to curves in 2-space.

The images were handcrafted with geogebra: https://www.geogebra.org/m/knptsh5m

The reference is at https://www.uoanbar.edu.iq/eStoreImages/Bank/4541.pdf

Answer 4

Intuitively,

You need a tension towards the center of curvature (if you image a weigtless rope connected to the particle instead of following a path). The same reason when you turn in a bike you mostly lean in the curve. Otherwise you would be flung out and not be turning the way you want.

It actually helps to think about it the other way. When a moving body has a perpendicular force applied (and hence acceleration) it forces the body to curve in the direction of the force. If that force is constant and remains perpendicular the resulting curve is a circle (think of a mary-go-round). If that force is constant, with also constant direction the resulting curve is a parabola (think gravity). An inverse square law force results in an ellipse (or hyperbola) for a curve (think planets and comets).

For the math see https://physics.stackexchange.com/a/83592/392

When a particle follows a path then at any instant there is tangential vector $\vec{e}$ and a normal vector $\vec{n}$ allowing for the velocity to be defined as $$\vec{v} = v \vec{e}$$ and the acceleration $$\vec{a} = \dot{v} \vec{e} + \frac{v^2}{\rho} \vec{n}$$ where $\rho$ is the radius of curvature of the path.

So your question is why does $\vec{a}_N =\frac{v^2}{\rho}\vec{n}$ always points towards the inside of the curve? In the post I show that for a function $y=f(x)$ the radius of curvature is

$$\rho(x) = \frac{\left(1+{f'(x)}^2\right)^\frac{3}{2}}{f''(x)}$$

and since the numerator is always positive, a positive radius of curvature corresponds to a positive 2nd derivative $f''(x)$. This happens to be up, for a cup shape ( $\cup$ ) and down for a cap shape ( $\cap$ ).

Answer 5

In order to move through a concave path, an agent has to impart force to otherwise a linearly-moving object. The object , by virtue of its motion, under the absence of any external force, always travels or tends to travel in the direction of the velocity vector at the concerned instant.

So, when the object has to transverse a curve trajectory, the main requisite is the introduction of a force that manipulates the direction of velocity such that the resultant locus is the required curvilinear path, otherwise the object would travel straight.

---

Now,the million-dollar question: what should be the direction of the force?

The direction of the force is evidently the direction of the acceleration or the limit of change of velocity with respect to time. So, to find the direction, let's think of an infinitesimal situation.

Let for a short amount of time $\Delta t$, the distance traveled is $v(t)\Delta t$ along a circular arc of radius $r$. The angle transversed is then $$\Delta\theta = \dfrac{v(t) \Delta t}{r}$$.

Imagine the bisector of the angle. Now, consider the changes in velocity parallel & perpendicular to this bisector. Initially the velocity has a component $v\sin(\frac{\Delta\theta}{2})$ away from the center & $v\cos(\frac{\Delta\theta}{2})$ transversely.Afterwards, it has a component $v\sin(\frac{\Delta\theta}{2})$ toward the center & $v\cos(\frac{\Delta\theta}{2})$ transversely as before. Thus the change of velocity is of magnitude $2v\sin(\frac{\Delta\theta}{2})$ toward the center of the arc.

As $\Delta\theta$ is vanishingly small, $\sin(\frac{\Delta\theta}{2})$ becomes indistinguishable as $\dfrac{\Delta\theta}{2}$. Thus, we can put $$|\Delta v(t)| = v^2 \dfrac{\Delta\theta}{r}$$. And the direction is towards the center. Thus the force is rotating the position vector along the curved trajectory and the change is radially inward regardless of whether it is traced clockwise or counter-clockwise.

---

The picture becomes more vivid if we calculate using polar co-ordinate.

First, we write the position vector as $\mathbf{r} = r\cdot e_r$ . Now consider the change of $\mathbf{r}$ with time. Its change during $\Delta t$ is $r\Delta\theta \cdot e_{\theta}$. $e_r \quad \& \quad e_{\theta}$ are mutually perpendicular, the first being outward radially from the center . Therefore velocity is $$v = \dfrac{d\mathbf{r}}{dt} = r\dfrac{d\theta}{dt}\cdot e_{\theta} = \omega r \cdot e_{\theta}$$.

By putting $r = 1$, we get $$\dfrac{d}{dt} (e_r) = \omega e_{\theta}$$.

Similarly, a change of $\theta$ implies a change of $e_{\theta}$.It can be seen that $$\dfrac{d}{dt} (e_{\theta}) = -\omega \cdot e_r$$. Now we differentiate the velocity, $$\mathbf{a} = \omega r \dfrac{d}{dt} (e_{\theta}) = -{\omega}^2 r\cdot e_r $$. . This result falls down automatically with the correct direction which is opposite of $e_r$ i.e. toward the center radially.