We need produce index.html self-contained. Need implement physics: heptagon regular, centered, spins 360° every 5 seconds. Balls 20 numbered 1-20 dropping from center affected by gravity and friction, collide with each other and bounce off rotating walls. Ball spin via rotation of number. Implement collision detection ourselves. No external resources.

We need carefully create code. Let's plan.

Canvas app. Use HTML5 canvas. Full window. Heptagon centered. Radius maybe 300. Rotates continuously. Balls spawn at center sequentially with small offset to prevent overlapping. Gravity downward (y positive). Friction: air resistance and maybe wall/ball restitution. Ball-ball collisions with impulse.

Need implement collision detection with rotating walls. The heptagon walls are line segments rotating about center. Each wall segment between vertices computed at current angle. Need detect collision circle-line segment. Since wall rotates, in world frame the wall moves; collision response should account for wall velocity at contact point due to rotation.

Simplify: In world coordinates, compute segment endpoints rotating with angular velocity ω = 2π/5 rad/s. For each ball, find closest point on each segment. If distance < ball radius, collision. Normal from closest point to ball center. Wall velocity at contact point = ω × (contact - center). Since 2D, v_wall = ω * perp(r) where perp(x,y)=(-y,x) times angular speed (rad/s) maybe sign. Angular velocity vector k; v = ω k × r = (-ω y, ω x). If rotating counterclockwise positive. Need consistent orientation.

Collision response: use impulse with relative velocity considering wall velocity. For static/moving wall, treat wall as infinite mass. Compute relative normal velocity vr_n = (v_ball - v_wall)·n. If penetrating and moving toward wall, apply impulse j = -(1+e) vr_n / (1/m) = -m*(1+e)*vr_n. But because wall velocity not static, should adjust. Actually separating impulse along normal for ball: v_ball' = v_ball + (j/m) n, where j solves (v_ball' - v_wall)·n = -e (v_ball - v_wall)·n. => j = -(1+e) vr_n * m. Good.

For ball-ball: standard 1D impulse along normal n from b1 to b2. equal mass m. v1' = v1 - j/m * n, v2' = v2 + j/m * n where j = -(1+e) vr_n / (1/m+1/m) = -m(1+e) vr_n/2. For equal mass.

Need friction between balls and walls? User asks gravity and friction. Could include angular damping (spin friction) and linear air friction. Maybe tangential friction impulse at collisions to affect spin and linear velocity. But spin asked as visual rotation of number; not necessarily physics spin (angular momentum). Simpler: each ball has orientation angle; angular velocity affected by collision tangential impulse / radius (roll/slip). Number rotation shows spin. Gravity/friction: apply linear drag each frame. Add rolling friction? Simpler: linear damping, angular damping.

But "show each ball's spin via the rotation of its number" implies each ball can spin; collisions may transfer spin. We can implement simple spin physics: each ball has angular velocity ω_b (radians/s). During collisions, tangential impulse due to friction modifies linear velocity tangential component and angular velocity. Use Coulomb friction with coefficient. For ball-ball/wall collisions, compute relative tangential velocity at contact including surface spins. For wall: relative tangential velocity = (v_ball - v_wall)·t - r*ω_b (with sign). Apply friction impulse limited. Update angular velocity: ω_b += (r * j_t / I) where I = 0.5 m r^2 for solid disk. For equal masses same radius. For ball-ball, relative tangential velocity at contact includes difference of linear tangential components and radii spins. Use j_t = -min(μ|j_n|, m_eff*|dv_t|/??) with tangential restitution maybe 0. We can implement standard friction impulse at collisions.

Given time, maybe simpler spin: give each ball initial random angular velocity; apply angular drag; collisions with wall/ball random angular impulse. But better include friction.

Need numerical integration: semi-implicit Euler. Fixed time step maybe 1/60 with sub-steps. Heptagon rotation angle updates.

Collision detection: circle-line segment. For each ball vs each of 7 walls. Also ball-ball O(n^2)=190 pairs, fine.

Need positional correction to prevent sinking/tunneling: after impulse, project ball out of wall by overlap along normal; for ball-ball, resolve overlap proportional to inverse mass (equal half). Maybe allow small overlap.

Wall endpoints: compute regular heptagon with radius R. But orientation: one vertex at top? Heptagon centered. For regular polygon with n=7, vertices at angle θ_i = baseAngle + 2π i/n. baseAngle maybe -π/2 so a flat edge at top? Center ball drop. We can set initial rotation so one side horizontal at top? Not critical. Let angle = current spin angle - π/2 + i*2π/7. Then radius maybe 280.

Spin: θ(t) = (2π/5)*t. Canvas coordinate y down.

Balls radius maybe 14-18. Number font.

Colors. Balls dropping from center: spawn at center with small random offset? If all at same point overlapping, they'll explode. Could spawn sequentially every 0.2s with slight offset. User says "20 numbered balls (1 to 20) dropping from the center". We can spawn them staggered from center with tiny upward push or random small offsets. Let's spawn all at start but with small random positions near center to avoid perfect overlap. Or at t=0, place them with slight jitter and zero initial velocity; gravity pulls them down. Since they overlap, resolver will separate. Better stagger spawn: every 0.25s spawn next ball at center (0,0) with slight initial random velocity. This avoids huge overlap.

Implement spawn timer. 20 balls. If spawn at exact center, radius overlapping only one ball at a time, ok. But then they accumulate at center; gravity pulls them. Heptagon centered. Fine.

Need ball numbering. Draw number rotated by ball.angle.

Let's design constants:
- canvas full window; resize.
- heptagonRadius = Math.min(width,height)*0.42 maybe.
- wall angular velocity = 2π/5 rad/s.
- gravity = 800 px/s^2 (down).
- ball radius = 16.
- ball mass = 1.
- moment of inertia I = 0.5*m*r*r.
- restitution e = 0.7 (bounce).
- friction coefficient μ = 0.3 maybe.
- linear damping = 0.999 per step or 0.995.
- angular damping = 0.99.
- dt = 1/60, substeps = 4 maybe.

Collisions per substep:
1. Update positions/angles: v += g*dt; v *= damping; p += v*dt; angle += ω*dt; ω *= angDamping.
2. Detect/resolve collisions iteratively maybe multiple iterations per substep.

Wall collisions: For each ball, for each wall segment:
- Compute endpoints A,B in world.
- Vector AB, AP.
- t = clamp((AP·AB)/(AB·AB),0,1).
- closest = A + AB*t.
- d = p - closest; dist = |d|.
- If dist < r and dist > epsilon:
   n = d/dist (from wall to ball).
   overlap = r - dist.
   v_wall = angularVelocityWall * perp(closest - center). Sign: spin angle increasing CCW. perp = (-y,x). Angular velocity scalar = +2π/5.
   rel = v - v_wall.
   vn = rel·n.
   if vn < 0:
     jn = -(1+e)*vn * m; // because denominator 1/m
     // friction
     tvec = perpendicular? Compute tangent unit vector. t = (-n.y, n.x) maybe orientation; but sign matters for spin. Define t such that cross(n,t) positive? Let's define t = (-n.y, n.x) (90° CCW). Relative tangential at surface: v_t = rel·t - r*ω_ball (positive direction t). Because point on ball due to spin = ω*r in t direction. We want impulse to reduce relative tangential velocity. 
     jt = -v_t * m? Effective mass for tangential? Since wall infinite. For friction impulse: desired Δv_t = -v_t. For ball, impulse along t changes v_t by jt/m and changes spin by -r*jt/I (because tangential force at contact causes torque). The surface relative tangential velocity change = Δv_t - r*Δω. Δω = (r*jt/I) with sign? Torque τ = r × F = r*jt (since F in t direction, lever r along -n? Actually contact point relative to center = -r n. Cross(-r n, jt t) = -r*jt*(n × t). If t = (-n_y,n_x), then n×t =? n=(nx,ny), t=(-ny,nx), cross(n,t)=nx*nx - ny*(-ny)=1. So torque positive. Angular acceleration positive. Surface velocity due to spin at contact = ω * k × (-r n) = -ω r t? Wait k×(-r n) = -r (k×n) = -r t (since t=k×n). So surface velocity = ω*(-r t) = -ω r t. So relative tangential velocity = (v - vwall)·t - r ω. Good. Impulse jt in +t on ball changes linear +jt/m t, and spin + r jt /I. Surface velocity change from spin = -r Δω t = -r*(r jt/I) t = -(r^2 jt/I) t. So total Δv_t_surface = jt/m - r^2 jt/I = jt*(1/m - r^2/I). For solid disk I=0.5 m r^2 => 1/m - 2/m = -1/m. So Δv_t = -jt/m. To cancel v_t, need jt = m*v_t. Coulomb limit: max |jt| = μ |jn|. So jt = clamp(m*v_t, -μ|jn|, μ|jn|). Sign: if v_t positive, jt positive reduces. Yes.
   Apply: v += (jn*n + jt*t)/m; ω += (r*jt)/I.
   Position correction: p += n*overlap.
- If dist == 0, normal ambiguous; skip or use outward normal from center? Could skip.

Ball-ball collisions:
- For each pair i<j:
   d = p2-p1; dist = |d|; if dist < 2r:
   n = d/dist (from 1 to 2).
   overlap = 2r - dist.
   // relative velocity at contact
   v_rel = v2 - v1.
   vn = v_rel·n.
   if vn < 0:
     jn = -(1+e)*vn / (1/m1+1/m2) = -m*(1+e)*vn/2 (equal).
     Apply normal impulse: v1 -= jn/m n; v2 += jn/m n.
     // friction/spin
     t = (-n.y,n.x).
     // relative tangential velocity at contact surface: (v2 - v1)·t - r(ω2 - ω1)? For ball2 surface velocity at contact = -r ω2 t; ball1 = -r ω1 t; relative tangential of ball2 wrt ball1 = (v2-v1)·t - r(ω2 - ω1). Wait surfaces due to spin: surface2 = -r ω2 t; surface1 = -r ω1 t; relative = (v2 - v1)·t + (surface2 - surface1)·t = (v2-v1)·t - r(ω2-ω1). Yes.
     v_t_rel = (v2 - v1)·t - r*(ω2 - ω1).
     Want reduce to 0. Effective mass for tangential impulse between two bodies with spin:
       Δv2_t = +jt/m; Δv1_t = -jt/m.
       Δω2 = +r*jt/I; Δω1 = -r*jt/I (impulse on ball2 in +t, on ball1 in -t; torque sign as before, for ball1 impulse -jt t at contact -r n: torque = (-r n)×(-jt t)= r jt (n×t)= r jt positive. So ω1 increases, Δω1 = +r jt/I. Wait check signs carefully. We want symmetry; friction force on ball2 in +t, on ball1 in -t. For ball1 contact point = +r n (since n points from 1 to 2, contact on ball1 is at +r n? Actually center1 to contact point is +r n. Then torque = r n × (-jt t) = -r jt (n×t) = -r jt. So ω1 decreases. Good. Need sign consistent with earlier wall case. For wall, contact point relative to ball = -r n (wall to ball). Fine. For ball2, contact point = -r n, force +jt t => torque +r jt. For ball1, contact point = +r n, force -jt t => torque -r jt. Thus Δω2 = +r jt/I, Δω1 = -r jt/I. Then surface velocity change for ball2 at contact from spin = -r Δω2 t = -r^2 jt/I t. For ball1 surface = -r Δω1 t = +r^2 jt/I t. Relative tangential surface change = Δ(v2-v1)_t + (surface2 - surface1)_t = (jt/m - (-jt/m)) - 2 r^2 jt/I = 2jt/m - 2r^2 jt/I = 2jt*(1/m - r^2/I) = for disk = 2jt*(1/m - 2/m)= -2jt/m. To cancel v_t_rel, need jt = m*v_t_rel/2? Since Δ = -2jt/m => set -2jt/m = -v_t_rel => jt = m*v_t_rel/2. Good. Limit |jt| ≤ μ |jn|. So jt = clamp(m*v_t_rel/2, -μ|jn|, μ|jn|).
   Position correction: p1 -= n*overlap*0.5; p2 += n*overlap*0.5.

Could perform collision iterations maybe 3 per substep to stabilize stacks. Since many balls, okay.

Heptagon drawing: fill maybe transparent stroke thick line. Draw rotating number label? Not required. Maybe draw walls as blue.

Balls draw: circle fill white with stroke; rotated number. Also maybe small dot to show spin.

Need ensure initial heptagon orientation and spin visible.

Potential tunneling at high speed; dt small, substep 4 maybe enough. Could also handle continuous wall with segment normals.

Let's code carefully.

Coordinate: canvas origin top-left; center = (cx,cy). World coordinates relative to center. y down. Gravity positive.

Compute vertices: radius R. angle = wallAngle - π/2 + i*2π/7. Then vertex v = (cx + R*cos(a), cy + R*sin(a)). Use world coords.

Wall angular velocity = 2π/5. Direction: angle increasing => CCW. perp = (-y,x).

For collision with rotating wall, contact point in world relative center = closest - center. v_wall = ω_wall * (-(closest.y-cy), (closest.x-cx)). Note in JS vector, use center.

Edge cases: If ball center is outside heptagon beyond a segment? With circle-line segment collision using finite segment covers interior. If ball escapes through vertex gaps? Since radius large enough and balls inside. However circle may be outside and closest point at vertex; segment check still handles vertices because t clamp to endpoints. If ball completely outside near a vertex, collision will push inward. Fine.

Spawn: start spawning after short delay. Use spawnInterval = 0.25s. ballCount = 0; timer accumulates dt. When timer > interval and ballCount<20, spawn. Spawn at center with radius=16. initial v=(Math.random()-0.5)*20, maybe -50 upward? Let's spawn at center (0,0) with v = ( (Math.random()-0.5)*50, (Math.random()-0.5)*50 ) to separate. Angular velocity random small.

UI: show count? Could show title/instructions. Maybe minimal: display "Balls: X/20" and FPS. Not required.

Let's write code. Need manage canvas scale for high DPI. Use ctx.setTransform(dpr,0,0,dpr,0,0); canvas width = window.innerWidth*dpr etc. But we can keep simple with css width/height and logical pixels. However for crisp text on high DPI, set canvas.width=innerWidth*dpr and scale. Let's implement.

Need variables:
let balls = [];
let wallAngle = 0;
let spawnIndex = 0;
let spawnTimer = 0;
const totalBalls = 20;
const SPAWN_INTERVAL = 0.25;

Loop:
function update(dt) {
  wallAngle += WALL_OMEGA * dt;
  spawnTimer += dt;
  if (spawnIndex < totalBalls && spawnTimer >= SPAWN_INTERVAL) {
    spawnBall(spawnIndex+1); spawnTimer=0; spawnIndex++;
  }
  const subSteps = 4;
  const sdt = dt/subSteps;
  for (let s=0; s<subSteps; s++) {
    integrate(sdt);
    resolveCollisions(sdt); // maybe iterate 3 times
  }
}

But spawnTimer resetting each spawn; if dt accumulated at 1/60, interval 0.25 => ~15 frames. Fine.

Better to spawn at start all? Stagger is nice.

integrate:
for ball: ball.vx += gx*sdt; ball.vy += gy*sdt; apply damping: ball.vx *= Math.pow(LIN_DAMP, sdt?); If damping factor per second, multiply by Math.exp(-linDrag*sdt) or just factor^sdt. Use pow for constant. Similarly angular. LIN_DAMP_PER_SEC = 0.99 maybe; factor = Math.pow(0.99, sdt). Positions update.

resolveCollisions: iterate maybe 3 times. Each iteration: wall collisions then ball-ball.

Potential double impulses causing energy blow. Use restitution 0.6, friction 0.2.

Ball numbers: draw text aligned center, baseline middle. Rotate by ball.angle. Because number orientation rotation; no need for rotation direction visualization beyond number.

Heptagon rendering: stroke width 8, color #4488ff.

Let's implement collision math functions.

Need efficient dot/cross. Write helper functions.

Let's consider collision resolution order and velocity of wall. Since wall infinite mass, normal impulse independent of mass denominator (m). For ball-ball equal mass.

Let's code wall collision carefully.

function resolveWall(ball, A, B, center, wOmega) {
  const ax = A.x, ay=A.y, bx=B.x, by=B.y;
  const abx=bx-ax, aby=by-ay;
  const apx=ball.x-ax, apy=ball.y-ay;
  const ab2=abx*abx+aby*aby;
  let t = 0;
  if (ab2 > 0) t = Math.max(0, Math.min(1, (apx*abx+apy*aby)/ab2));
  const cx = ax + abx*t, cy = ay + aby*t;
  const dx = ball.x - cx, dy = ball.y - cy;
  const dist2 = dx*dx+dy*dy;
  const r = ball.r;
  if (dist2 >= r*r || dist2 === 0) return;
  const dist = Math.sqrt(dist2);
  const nx = dx/dist, ny = dy/dist;
  const overlap = r - dist;
  // wall velocity at contact
  const rx = cx - center.x, ry = cy - center.y;
  const vwx = -wOmega * ry, vwy = wOmega * rx;
  // relative velocity of ball wrt wall
  const rvx = ball.vx - vwx, rvy = ball.vy - vwy;
  const vn = rvx*nx + rvy*ny;
  if (vn < 0) {
    const jn = -ball.m * (1 + RESTITUTION) * vn;
    // tangent
    const tx = -ny, ty = nx;
    const vt = rvx*tx + rvy*ty - r*ball.omega;
    // effective mass for tangential
    // jt needed to zero vt = ball.m * vt (derived)
    let jtRaw = ball.m * vt;
    const maxF = FRICTION * Math.abs(jn);
    if (jtRaw > maxF) jtRaw = maxF;
    else if (jtRaw < -maxF) jtRaw = -maxF;
    const jtx = jtRaw*tx, jty = jtRaw*ty;
    ball.vx += (jn*nx + jtx)/ball.m;
    ball.vy += (jn*ny + jty)/ball.m;
    ball.omega += (ball.r * jtRaw) / ball.I;
  }
  // position correction
  ball.x += nx * overlap;
  ball.y += ny * overlap;
}

Wait derived jtRaw = m*vt. Let's verify sign. We have Δvt_surface = -jt/m. We need Δvt_surface = -vt (to zero). So -jt/m = -vt => jt = m*vt. If vt positive, jt positive. That matches friction force on ball in t direction to accelerate surface to reduce vt. Good.

But note friction impulse sign maybe opposite because vt definition includes -rω. We'll trust.

For ball-ball:

function resolveBallBall(a,b) {
  const dx = b.x - a.x, dy = b.y - a.y;
  const dist2 = dx*dx+dy*dy;
  const minD = a.r + b.r;
  if (dist2 >= minD*minD || dist2 === 0) return;
  const dist = Math.sqrt(dist2);
  const nx = dx/dist, ny = dy/dist;
  const overlap = minD - dist;
  const rvx = b.vx - a.vx, rvy = b.vy - a.vy;
  const vn = rvx*nx + rvy*ny;
  if (vn < 0) {
    const jn = -(1+RESTITUTION)*vn / (1/a.m + 1/b.m);
    a.vx -= jn/a.m * nx; a.vy -= jn/a.m * ny;
    b.vx += jn/b.m * nx; b.vy += jn/b.m * ny;
    // friction
    const tx = -ny, ty = nx;
    const vt = rvx*tx + rvy*ty - a.r*(b.omega - a.omega); // equal radii
    // jt to zero = m*vt/2 for equal mass, else general formula
    // General effective: for tangential impulse jt on b (+t) and -jt on a, surface rel change = jt*(1/ma+1/mb) - ra^2/Ia * jt - rb^2/Ib * jt? Need derive signs. For ball b contact lever = -rb n, force +jt t => torque +rb jt; Δωb=+rb jt/Ib; surface b change = -rb Δωb = -rb^2 jt/Ib. For ball a contact lever = +ra n, force -jt t => torque = ra n × (-jt t) = -ra jt (n×t) = -ra jt; Δωa = -ra jt/Ia; surface a change = -ra Δωa = +ra^2 jt/Ia. Relative surface change = Δ(vb-va)_t + (surf_b - surf_a) = jt(1/ma+1/mb) - rb^2/Ib * jt - ra^2/Ia * jt. Set = -vt. So jt = -vt / (1/ma+1/mb - ra^2/Ia - rb^2/Ib). For disk I=0.5mr^2 => denominator = 1/m+1/m - 2/m -2/m = -2/m. jt = -vt / (-2/m) = m vt/2. Good.
    let denom = (1/a.m + 1/b.m) - (a.r*a.r)/a.I - (b.r*b.r)/b.I;
    if (Math.abs(denom) < 1e-9) denom = 1e-9;
    let jtRaw = -vt / denom;
    const maxF = FRICTION * Math.abs(jn);
    if (jtRaw > maxF) jtRaw = maxF;
    else if (jtRaw < -maxF) jtRaw = -maxF;
    a.vx -= jtRaw/a.m * tx; a.vy -= jtRaw/a.m * ty;
    b.vx += jtRaw/b.m * tx; b.vy += jtRaw/b.m * ty;
    a.omega -= (a.r * jtRaw)/a.I;
    b.omega += (b.r * jtRaw)/b.I;
  }
  // position correction
  a.x -= nx*overlap*0.5; a.y -= ny*overlap*0.5;
  b.x += nx*overlap*0.5; b.y += ny*overlap*0.5;
}

Check signs of angular updates for ball-ball: For ball b force +jt t at contact point -r n => torque +r jt (n×t). n×t = 1. So Δωb = +r jt/I. For ball a force -jt t at contact +r n => torque = -r jt. So Δωa = -r jt/I. Good.

Now about wall angular velocity direction and normal orientation. For a rotating wall inside heptagon, if ball inside and wall rotates, collision normal from closest point to ball center points inward. v_wall computed. Should produce realistic bounce with moving wall. Test mentally: ball at rest near right wall, wall rotating CCW, contact point on right side moving upward? At right side (x positive), v_wall = (-ω y, ω x). For y=0, v = (0, +ωx) upward. Ball above wall moving up; relative velocity ball - wall = (0 - 0, 0 - ωx) = downward; vn negative (moving toward wall), impulse positive normal and perhaps ball gets upward velocity after bounce? Let's compute: n points right from wall to ball. rel = (0,-vwall_y). vn = -vwall_y (negative). jn = -m(1+e)vn = positive (1+e)*vwall_y. v += jn/m n = positive x? Wait n = (1,0). That pushes ball right, not upward. Hmm not capturing tangential wall motion. Because the normal impulse only handles normal relative velocity, but relative tangential velocity (wall moving up) is in t direction (up), handled by friction. Normal relative velocity due to wall moving toward ball is along normal? At rightmost flat vertical wall, CCW rotation means wall moves upward, not into ball. If ball is inside and wall moves tangentially, normal relative velocity zero; collision won't push ball upward unless friction transfers tangential motion. Realistically a moving vertical wall should drag ball upward via friction. Our friction impulse handles tangential relative velocity: vt = rel·t - rω. For t = (-ny,nx) = (0,1). rel·t = -vwall_y. vt negative (assuming wall up). jtRaw=m*vt negative. Max μ|jn| = 0 (since vn=0, no normal impulse, so no friction! Coulomb friction requires normal force). Since no normal contact force, no friction, so ball unaffected. That matches physical: if wall just slides tangentially without pressing, there's normal contact but zero normal relative velocity; static friction can still act if not sliding, but kinetic friction requires normal force; with no normal force, no friction. In reality a wall moving tangentially while maintaining contact exerts kinetic friction because there is contact force (gravity or centripetal). But our impulse-based collision only generates normal impulse when approaching; during persistent contact with zero normal relative velocity, no friction. Could add resting contact friction using gravity normal force, but complex. The user probably expects balls bounce off rotating walls; normal impulse from wall moving toward/away from ball will be handled. For tangential motion, maybe friction not necessary; but we can still include friction when normal impulse occurs (bounces). Good enough.

However if wall rotates, for a ball resting on a slanted wall, gravity pulls it into wall; normal relative velocity due to wall motion normal component can create impulses. Works.

But need handling of resting contacts to prevent sinking: position correction pushes out; and friction during collisions adds spin.

Maybe implement a simple resting friction/rolling for balls inside heptagon? Not necessary.

One issue: balls may pass through walls at high relative tangential speed because position correction only pushes along normal; if ball at wall moving parallel, it remains in contact due to correction; next frame, if no normal impulse, it can drift through. But heptagon is convex; gravity keeps them inside; if spin flings them, okay.

Potential problem: ball colliding with vertex: closest point endpoint. v_wall computed at endpoint. Good.

Need ensure balls stay within heptagon after spawn. At center, gravity pulls down; they drop and hit bottom. Good.

Need display spin of number. ball.angle orientation. Angular velocity from collisions and initial random. Also because ball rotates as it rolls along wall? We implement spin dynamics through friction; but a ball sliding down a frictionless wall wouldn't spin. Friction tangential impulse during bounce and collisions will spin them. We might also add a rolling constraint? Not needed.

Maybe to make spin visible, add initial random angular velocities and color markers? Number rotation enough.

Let's consider numerical stability. Many balls in confined space, lots of collisions. Restitution 0.5 might reduce jitter. Friction 0.2. Damping. Use substeps 5. Maybe fine.

Let's implement spawn: ball object:
{
 id, x,y, vx,vy, angle, omega, r, m, I, color
}
Spawn at (0,0) relative center. initial small random velocity. angle random. omega random between -5 and 5.

Colors: use HSL hue = i*18.

Drawing: For each ball, translate to center+ball.x, rotate ball.angle, draw circle, then number. Use ctx.arc. Fill with slight radial gradient? Simple.

Number text: fill black. Font bold 14px sans-serif. Rotate by ball.angle. Since canvas y down, positive rotation is clockwise. We display number orientation; direction of spin visible. Fine.

Heptagon fill maybe none, stroke. Background dark.

Let's also show small status text: "Heptagon spin: 360° / 5 s" and balls count. Place at top-left. But canvas transforms; draw before transforms.

Potential issue: high DPI scaling and text font. We'll implement.

Let's code full HTML. Use requestAnimationFrame with time delta. Use dt = (timestamp - last)/1000; clamp dt.

Need set canvas size on resize.

Let's write.

But before final, test mentally. We should maybe run in head? No actual browser in environment. But we can verify syntax by maybe node? Not necessary; but we can use bash to check no external files.

Let's write index.html. Ensure no external resources. Use inline style/script.

Need include a note about sandbox? Not necessary.

Let's produce code with comments explaining custom collision detection.

One more thought: wall-ball position correction can push ball through opposite wall if overlap big; but substep small. Fine.

Maybe handle collision detection with rotating walls in local frame? We did world frame.

Let's implement vector helpers inside script.

Detailed update loop:

const WALL_OMEGA = 2*Math.PI/5; // rad/s
const SUBSTEPS = 5;

function update(dt) {
  wallAngle += WALL_OMEGA * dt;
  spawnTimer += dt;
  if (spawnIndex < TOTAL && spawnTimer >= SPAWN_INTERVAL) { spawn(); spawnTimer=0; }
  const sdt = dt/SUBSTEPS;
  for (let i=0;i<SUBSTEPS;i++) {
    integrate(sdt);
    for (let iter=0; iter<3; iter++) {
      resolveWallCollisions();
      resolveBallCollisions();
    }
  }
}

But spawnTimer resetting to 0 each spawn may cause slight average interval exactly SPAWN_INTERVAL if dt ~0.016; after 15 frames triggers. Fine.

Integrate:
function integrate(sdt) {
  const lin = Math.pow(LIN_DAMP, sdt);
  const ang = Math.pow(ANG_DAMP, sdt);
  for ball:
    ball.vx += GRAVITY_X*sdt;
    ball.vy += GRAVITY_Y*sdt;
    ball.vx *= lin; ball.vy *= lin;
    ball.x += ball.vx*sdt;
    ball.y += ball.vy*sdt;
    ball.omega *= ang;
    ball.angle += ball.omega*sdt;
}

Note gravity constant.

Resting contacts: Could add a small gravity-based wall push? Not.

Let's define constants:
const BALL_RADIUS = 16;
const RESTITUTION = 0.55;
const FRICTION = 0.25;
const LIN_DAMP = 0.999; // per second? Actually pow with sdt. Maybe 0.99.
const ANG_DAMP = 0.995;

Need heptagon radius: maybe Math.min(w,h)*0.42. With ball radius 16, 20 balls area maybe 16000 px; heptagon area ~3.6339 R^2. For R=250 area ~227k, plenty.

Spawn initial velocity: (Math.random()-0.5)*60, maybe downward? To drop. Let's set vy = Math.random()*50 (down) or 0. Actually from center, gravity will accelerate. Slight random.

Let's ensure balls don't spawn inside existing ones; since at exact center, first ball at center, next at center overlapping. That will cause strong separation. Better stagger and spawn at random small offset not overlapping. We can spawn at (Math.cos(i*1.3)*2, Math.sin(i*1.3)*2) with r=16; overlapping only if exactly same. Since radius same, two balls at center overlap fully. The first ball may have moved by next spawn if substep; at dt ~0.016, not much. So we can spawn with small offset: x = (Math.random()-0.5)*BALL_RADIUS*0.5 etc. Good.

Number rendering: Because ball rotates, number may be upside-down. Fine.

Let's consider ball angle unbounded; can normalize but not necessary. Use ctx.rotate(ball.angle). Text drawing: ctx.fillText(String(ball.id),0,0). To keep number upright? The request says show spin via rotation, so rotated.

Let's write complete file. Then final message concise.

Before writing, we can maybe mention current date? Not needed.

Need ensure no use of external images. Fine.

Let's produce.

The file is written. Need final concise response. Also maybe verify file exists. Use bash ls? Not necessary, but can confirm. The instruction says end with index.html present; we can list.