Mathematical proof for: Acceleration always orthogonal to its velocity changes its direction

Question

Whenever I ask this doubt that how can force perpendicular to objects velocity can change its direction but cant change its magnitude I get proof for only the magnitude will be constant that is: no component of force is parallel to velocity so it cant change the magnitude of velocity but I cant get proof for how can it change direction of motion. Every time they say that since force cant change magnitude it changes its direction. But I need proof of how orthogonal force change the direction of velocity.

edit-My problem is in my class my teacher said if acceleration is always perpendicular velocity then the magnitude of velocity will never change and proved it like this: $$\frac{d(v^2)}{dt} =\frac{d(v⃗.v⃗)}{dt}=a⃗ ⋅v⃗ =0$$ (since $a⃗$ and $v⃗$ are perpendicular to each other) So $v$ (speed=magnitude of velocity) is constant. After that he said since this acceleration cannot change magnitude of velocity so it will change only direction of velocity and said in general it is valid for any vector and said if a vector and its derivative are perpendicular to each other then only direction of vector will change and magnitude of that vector will remain constant. But I cannot understand how can perpendicular acceleration change the direction of velocity, I searched for the answer in the internet for hours and cant get a satisfactory answer I only get the same proof(that magnitude will remain constant) but in different different ways. Please if anyone can give a good explanation on how it changes direction of velocity(without changing its magnitude).

Answer 1

If a vector $\mathbf v$ is changing (so $\mathbf a = \frac{d}{dt} \mathbf v \neq 0$) but its magnitude is constant ($\frac{d}{dt}|\mathbf v|=0$), then the only thing that can be changing is its direction.

If you wish, you can write a 2-dimensional vector as $$\mathbf v = v \pmatrix{\cos(\theta)\\\sin(\theta)}$$ where $v\equiv |\mathbf v|$ and $\theta$ specifies its orientation with respect to the horizontal axis. From there, $$\frac{d}{dt}\mathbf v = \frac{dv}{dt} \pmatrix{\cos(\theta)\\\sin(\theta)} + v \frac{d\theta}{dt}\pmatrix{-\sin(\theta)\\\cos(\theta)} \equiv \mathbf a_\parallel + \mathbf a_\perp$$

It's easy to check that $\mathbf a_\parallel$ is parallel to $\mathbf v$, while $\mathbf a_\perp$ is perpendicular to $\mathbf v$. If the magnitude of the vector is constant - so $\frac{dv}{dt}=0$ - then only the second term survives, and is proportional to the rate of change of $\theta$. Therefore, $\mathbf a_\perp$ corresponds to a change in the direction of $\mathbf v$.

The extension to 3D is straightforward, and can be explicitly checked by writing $$\mathbf v= v\pmatrix{\sin(\theta)\cos(\phi)\\\sin(\theta)\sin(\phi)\\\cos(\theta)}$$ where now $\theta$ and $\phi$ are the polar and azimuthal angles, respectively.

Answer 2

As you stated, a force acting perpendicular to the velocity of a particle can't change its magnitude, as the component of the force acting along the direction of instantaneous velocity is zero.

The fact that such a force changes the direction of the particle's motion is simply because even though none of the force is increasing the particle's speed along its instantaneous direction of motion, ALL of the force is increasing the particle's speed along the direction of the force, i.e., its instantaneous acceleration is in the direction of the force.

Since this acceleration is not parallel to the particle's motion(it is gaining velocity in the direction of acceleration, that's what it means to accelerate), we can safely conclude that the particle must be changing its direction.

Similarly, for a force that's always perpendicular to a particle's velocity, it will, at all instants, only change the particle's direction of motion(towards the direction of the force), without changing its magnitude; an example being a central force, responsible for uniform circular motion.

@J.Murray has given an appropriate mathematical proof for the same.

Answer 3

Acceleration, $\vec{a}=\frac{d\vec{v}}{dt}$. Let's assume $\vec{a} \perp \vec{v}$ at all times and $\vec{a}\neq \vec{0}$.

$\vec{a} \perp \vec{v} \Rightarrow \vec{v}\cdot \vec{a}=0$. i.e, $\vec{v}.\vec{a}=\vec{v}\cdot \frac{d\vec{v}}{dt}=0 \quad ...(\star)$

Consider: $\frac{dv^2}{dt}=\frac{d\vec{v}\cdot \vec{v}}{dt}=\vec{v}\cdot \left(\frac{d\vec{v}}{dt}\right)+\left(\frac{d\vec{v}}{dt}\right)\cdot\vec{v}=2\vec{v}\cdot\left(\frac{d\vec{v}}{dt}\right)$, using this in $(\star) $, we get:

$\quad \vec{v}.\vec{a}=\vec{v}\cdot \frac{d\vec{v}}{dt}=\frac{1}{2}\frac{dv^2}{dt}=0 \Rightarrow v^2=constant \quad ...(\star\star)$.

$\therefore$ whenever a non-zero acceleration acts perpendicular to the velocity at all times, magnitude of velocity stays constant. Now read the sequence of equations ($\star\star$) from right to left. It says: if the magnitude of velocity is a non-zero constant then dot product of acceleration and velocity is zero $\Rightarrow$ either $\vec{a}=\vec{0}$ or $\vec{a}\perp \vec{v}$ at all times.

Answer 4

Mathematically, acceleration is defined as, $$\vec{a}=\frac{d\vec{v}}{dt}$$ Hence, we can define change in velocity as $$\vec{v}(t+dt)-\vec{v}(t)=d\vec{v}=\vec{a}dt$$ $$\vec{v}(t+dt)=\vec{v}(t)+\vec{a}dt$$ Now, if initial velocity, $\vec{v}(t)\perp\vec{a}$, you would be able to show that final velocity, $\vec{v}(t+dt)$ is not parallel to the initial velocity by the simple vector addition in last equation. Hence, direction of velocity is changed by an acceleration perpendicular to it.