Why does friction act towards the center if there is no relative motion in that direction?

Dynamic friction opposes relative motion. Static friction doesn't oppose motion itself, it opposes motion the coin would experience if there was no friction.

Static friction is the opposite to the net force the coin would experience without friction, up to some maximum value. If the net force (without friction) reaches that value, the static friction won't be able to counteract other forces, the coin will start moving and the friction will be dynamic friction.

At least in 101 physics courses, we often model dynamic friction by $$mathbf{F}_{text{dynamic}} = mu_{text{D}} ||mathbf{N}||$$ where $mathbf{N}$ is the normal force and $mu_{D}$ is the dynamic friction coefficient (determined experimentally). Note that dynamic friction does not depend on the speed. The maximum for static friction has an analogous expression $$mathbf{F}_{text{static (max.)}} = mu_{text{S}} ||mathbf{N}||text{.}$$

When you say this sentence you are very correct:

I know friction opposes relative motion

But remember that you can oppose relative motion in two ways:

• You can oppose it by trying to stop it.
• You can oppose it by trying to prevent it.

In the former case we call it kinetic friction. In the latter case we call it static friction.

It is the latter case with static friction you are experiencing in your scenario. Think of when you are pushing on a sofa but can't move it. It is static friction which is preventing you from initiating sliding (relative motion).

So, in conclusion, opposing relative motion does not require that a relative motion is already happening.

Static friction must point in the opposite direction to the relative motion that would occur in its absence. The confusion arises when trying to find the instantaneous relative velocity between the coin and the surface if friction were to disappear. Here's a quick mathematical proof that might help understand what's happening:

If friction were to suddenly disappear, the coin would move along the tangent, and the disk would rotate some angle $Deltatheta$.

$$vec{v}_A text{ (velocity of coin)} = romegahat{i} vec{v}_B text{ (velocity of surface)} = romegacosDeltatheta hat{i} - romegasinDeltatheta hat{j} vec{v}_{AB} = romegaleft[(1-cosDeltatheta)hat{i} + sinDeltathetahat{j}right] $$

$$ left|vec{v}_{AB}right| = romegasqrt{1+cos^2Deltatheta - 2cosDeltatheta+sin^2Deltatheta}=romegasqrt{2-2cosDeltatheta} $$ $$ tanalpha = frac{sinDeltatheta}{1-cosDeltatheta}=frac{2sinfrac{Deltatheta}{2}cosfrac{Deltatheta}{2}}{1-1+2sin^2frac{Deltatheta}{2}}=cotfrac{Deltatheta}{2} alpha = tan^{-1}left(cotfrac{Deltatheta}{2}right)=tan^{-1}left(tanleft(90-frac{Deltatheta}{2}right)right)=90-frac{Deltatheta}{2} $$

$$ text{magnitude of } vec{v}_{AB} = romegasqrt{2-2cosfrac{Deltatheta}{2}},qquad text{direction of } vec{v}_{AB} = alpha = 90 - frac{Deltatheta}{2} lim_{Deltathetato 0} left|vec{v}_{AB}right| = romegasqrt{2-2cos0}=0,qquad lim_{Deltathetato 0} alpha = 90 - frac{0}{2} = 90 $$

Here, $alpha$ is the angle between the relative velocity vector and the x-axis. Now, to find the instantaneous relative velocity, you take the limits of both the magnitude and direction as $Delta theta$ approaches $0$. As expected, the magnitude goes to $0$, however, surprisingly, the angle goes to $90$ i.e the vector points radially outwards. Ergo, static friction points radially inwards and opposes relative motion.

I know friction opposes relative motion or if surfaces tend to have relative motions but there is no relative motion between the coin and the contact point on the disc in radially outward direction.

The reason there's no relative motion is because of static friction!

I'd first suggest a little experiment.

Take an empty transparent bottle (plastic water/soda bottle will do). Take a coin or a small object like a marble and put it in the bottle. Put the cap on. Holding the bottle by the cap, spin it around (you can spin yourself around). If you do this right, you will notice that the small object migrates outwards. Even if you slant the bottle such that the object has to move upwards, it will still work.

Here is the experiment setup:

Next, check out this animation I made: https://www.desmos.com/calculator/s8roerbvub. Slide/play the value of $t_0$ to analyze the motion of a particle moving at $1 rm m/s$ in a straight line and in circular motion.

Hopefully this builds your intuition that without some center seeking force, the object will go outwards.

Now, time for some reasoning.

From the animation, you can see that the particle "wants" to go outwards. At $t_0=0$, the particle is instantaneously following the circular trajectory of a point on the rotating disk, and so there is no relative motion. But the particle will try to push out, so friction will push inwards to resist the relative motion.

Why is friction pointing directly inwards? Because there is no angular acceleration. If the disk accelerates, then the friction force will not be directed towards the center -- a component of it will speed the particle up.

Nonetheless, in the case where the angular speed is constant, why does friction point directly inward?

First, consider the following sketch:

It shows a particle moving in uniform circular motion with constant speed, and geometrically you can see that on some "small" time interval $delta t$, friction will always point inwards.

Second, we know that if the speed of the particle is to remain constant, force must always be perpendicular to trajectory. Therefore, there is nowhere else for friction to point except inwards.

Friction pointing directly inwards will occur if the particle and disk are moving at the same uniform angular speed. The particle will want to go directly outwards, so friction will want to resist that motion. The only geometric way to resist that motion is for the friction force to point directly inwards.