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0:00
Welcome back, this is Amit Sinha. Today, projectiles, and the polar view of motion. Motion in a plane, done the elegant way, with geometry doing most of the heavy lifting. And when the ground itself starts spinning, we will build coordinates that spin along with it, and watch Newton's law grow four new teeth.
0:24
Two identical bullets, held at exactly the same height above a level field. One is fired perfectly horizontally, at nine hundred meters per second. The other is simply dropped from beside the muzzle, at the same instant. One screams across the field for nearly a kilometer. The other lands at your feet. Here is the question. Which of them hits the ground first? They hit together. Every single time. Gravity runs one clock, and that clock does not care how fast you cross the field while it ticks.
1:01
The fired bullet is doing two things at once. It coasts sideways at a constant nine hundred, and it falls, exactly like its dropped twin, matching it centimeter for centimeter on the way down. Two separate stories, sharing one body. That is the whole secret of today, and it is worth saying slowly. Split the motion into independent stories. Solve each story alone, with tools you already own. Then add the stories back, like vectors. That one idea will knock down a hillside cannon,
1:40
fence off the sky, and save or doom a falling monkey. Here is the plan. First, componentwise motion, with gravity writing only the vertical story. Second, the trajectory equation, the curve left behind when you fire the clock. Third, three classic problems that need nothing but those two ideas. And fourth, polar coordinates, the frame that rides along with a turning motion, and the four terms it hands us.
2:12
Start where every dynamics problem starts, with the acceleration. Horizontally there is no force, so x-double-dot is zero. Vertically, gravity writes y-double-dot equals minus G. Now give the launch its proper name. The initial velocity v_0 leaves at angle theta, and resolves into two components, v_0 cos theta sideways, v_0 sine theta upward. Two clean numbers, and the problem splits in half. Lets integrate the horizontal story once, acceleration to velocity.
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x-double-dot is zero, so v_x never changes from its birth value, v_0 cos theta. Watch the dot run the X axis alone. Tick after tick, the same stride. No force, no change, inertia keeping perfect time. The vertical story integrates just as honestly. v_y starts at v_0 sine theta, and gravity docks it G every second. The slide keeps the running balance. The dot on the Y axis climbs, slows, pauses, and falls back. Now run both stories together.
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Their crossing point is the projectile, and the parabola blooms out of two straight-line motions. How long does the flight last? Ask the vertical story alone. The climb ends when v_y reaches zero, at time v_0 sine theta over G. The fall retraces the climb exactly, the same time down as up. Double it, and the whole flight lasts two v_0 sine theta over G. Watch the Y dot bounce. Its clock is set by the launch alone.
4:09
Integrate once more, velocity to position. The horizontal story gives X equals v_0 cos theta, times T. The vertical story gives Y equals v_0 sine theta, times T, minus half G T squared. Position fully solved. But both answers still wear the clock, and the curve in the sky does not. So fire the middleman. Solve the horizontal story for T, which is X over v_0 cos theta, and feed it to the vertical one. What remains is the trajectory equation, pure geometry. The height is X tan theta, minus G X
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squared over two v_0 squared cos squared theta. One curve, no schedule. Sweep the launch angle, in pairs. Twenty five degrees with sixty five. Thirty five with fifty five. Each pair lands on the very same spot, a flat twin and a lofted twin for every target. And between them all, the forty five degree curve reaches farthest, the champion of the family. Now look at the fan's outer edge. The shots crowd against a skyline that none of them ever cross, and its shape looks suspiciously like
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a parabola itself. Hold that suspicion. Two problems from now we will earn its equation, and fence the whole sky with it.
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The throw's most famous number, the range. Set Y to zero, keep the nonzero landing time, and X becomes two v_0 squared over G, times sine theta cos theta. The double angle identity folds that product into one clean function, and the range is v_0 squared sine two theta, all over G. Watch one shot fly, and read R straight off the ground. Which angle throws farthest? Differentiate the range with respect to theta. The derivative rides cos of two theta, and it vanishes when two theta reaches ninety.
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Forty five degrees, now proven rather than guessed, and the maximum range is v_0 squared over G. The fan agrees. Every neighbor of forty five lands shorter, on either side.
6:38
Time to test these formulas on something alive. A hunter aims a tranquilizer dart at a monkey hanging from a branch, thirty meters out and ten meters up. The monkey has studied physics, or thinks it has. At the muzzle flash, it lets go of the branch, planning to drop safely under the shot. Where should the hunter aim? Below the branch, to lead the falling target? Every instinct in your body wants a correction. Turn gravity off and the answer becomes obvious.
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With no gravity, the dart flies a perfectly straight line up the aim line. And with no gravity, the monkey lets go and simply hangs there, frozen at the branch. So aim straight at it, and the gravity-free story ends in a certain hit. Now remember superposition. The real story is exactly that one, plus falling. Switch gravity back on, and both bodies fall. They fall identically, half G T squared below their gravity-free ghosts, whatever their other motion happens to be.
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A thirty meter per second dart arrives in just over a second, sagging five and a half meters below the aim line. And the monkey has dropped the same five and a half meters below the branch. Equal sag, and the collision survives gravity untouched. So the hunter aims straight at the monkey, and letting go was the monkey's fatal mistake. Take the quick check while it is fresh. Send gravity to zero, and both ghosts become the whole story. The dart sails the aim line home, superposition with
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nothing left to add. Faster darts just meet the monkey higher up. No calculus showed up for work today.
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A cannon on a hillside. The ground climbs away at angle beta, and we fire up the slope at speed v_0, choosing launch angle theta. On flat ground the best launch is forty five degrees, and we proved that two slides ago. On a slope, forty five must waste range, but which way should we lean, and by how much? Start from what we know, and tilt the ground. On flat ground the champion at forty five bisects the ground and the vertical. Watch what happens if we simply tilt the ground
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beneath the same fan of shots. The old winner is no longer the bisector. The shot that now splits the angle between the tilted ground and the vertical is the one at sixty degrees. That is the whole insight, and it holds at every slope angle. The best launch bisects the slope and the vertical. theta star equals forty five degrees, plus half of beta. On a thirty degree hill, sixty degrees. And on flat ground, beta of zero, we recover the plain forty five. That famous angle was never magic.
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It was a bisection all along. The bookkeeping confirms the picture. Tilt the axes to the slope, let gravity split into its along-slope and across-slope parts, and demand the shot land back on the hill. Maximize, and the greatest range along the slope comes out as v_0 squared over G, divided by one plus sine beta. Send beta to zero and the flat classic returns intact. Now swing the same cannon around and fire down the slope instead. Watch the ground sweep below level.
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The geometry flips the sign of beta, and the divisor becomes one minus sine beta. On a thirty degree grade the ratio of the two ranges is three, exactly. Same cannon, same powder, three times the reach, purely from choosing which way the hill runs.
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One cannon, one muzzle speed, every possible angle, firing forever. Some of the sky gets hit constantly, some of it never. What exactly can this cannon reach, and where is it forever safe to stand? The fan piles up against that skyline we glimpsed earlier. Today we earn its equation, and the technique is worth more than the answer. The move is a shift of viewpoint. Stop following shots outward. Pick a target, an X and a Y, and ask which launch angles reach it.
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One identity rewrites the cos in terms of tan, so the whole equation now speaks tan theta. And what stands there is a quadratic in tan theta, with the target's coordinates as its coefficients. Now just read the roots. Two real solutions mean two shots reach the target, our flat and lofted twins. One solution means a single grazing shot. No real solutions mean the target is untouchable at this muzzle speed. The whole map of the sky is written in a discriminant.
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So set the discriminant to zero and see where the answers run out. Write B squared equals four A C with our coefficients, on the slide, step by step. The X squared factors cancel, and solving for the height gives the fence. Y equals v_0 squared over two G, minus G X squared over two v_0 squared. A single parabola, opening downward, crowning directly overhead. Read it like a map. Below the fence, two angles reach every target. On the fence, the twins merge into one grazing
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shot, tangent to the boundary. And the quick check is right overhead. Set X to zero and the ceiling reads v_0 squared over two G, exactly the height of a shot fired straight up. Artillery officers called that curve the parabola of safety. The lesson outlives the cannon. Boundaries do not come from chasing each trajectory to its personal maximum. They come from tangency, from asking when solutions appear and when they vanish. In the energy lectures the same discriminant thinking will
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decide which orbits can exist at all.
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Now stand up from the page. A real cannon owns every direction, not only one vertical plane. Take the parabola of safety and spin it about the vertical. The fence becomes a shell, a dome of safety. Under the dome, every point can be struck. Outside it, nothing ever will be. The gold circle on the ground marks the farthest reach, radius v_0 squared over G, the maximum range aimed in every direction at once. And the crown overhead is the straight-up ceiling.
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In the corners, the four numbers a projectile owns. Range, flight time, ceiling, and fence. One flat page equation, standing guard over a whole hemisphere of sky.
14:00
From the free sky to a leash. The next family of problems moves on circles, so we need the one number circular motion charges, the inward acceleration. No calculus needed, just a picture. Watch a particle in uniform circular motion, and freeze it at two nearby instants. The radius sweeps through a small angle. And the velocity, always at a right angle to the radius, turns through exactly the same angle. Now build the two triangles of change. The positions R one and R two differ by
14:38
a small step between them. The velocities V one and V two, drawn side by side, differ by their own small step. Same apex angle in each, both isosceles, so the two triangles are similar. And similar triangles trade in equal ratios. The velocity step over V matches the position step over R. Divide both sides by the little time interval, and let it shrink. The velocity step per time becomes the acceleration. The position step per time becomes the speed.
15:15
Cross multiply, and the inward acceleration is V squared over R. Prefer the spin rate? V is omega R, so the same push reads omega squared R. Keep both forms, because problems hand you either one. Hold on to that inward charge, it anchors everything ahead. But the circle is the tame case. The radius never changes, only the angle sweeps. What happens when radius and angle both change at once? For that we need coordinates built for turning, and that is where we go next.
15:55
Time to build those turning coordinates. Cartesian axes are honest workers, but they sit still. Make the motion circular, and X and Y tangle into sines and cosines that hide the simplicity underneath. So we describe a point by a distance from center and an angle of sweep. X is R cos theta, and Y is R sine theta. Now plant the basis on the moving point itself. r-hat is the unit vector pointing radially outward, along the line from the origin through the point.
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theta-hat is the unit vector pointing counterclockwise, tangentially, always at a right angle to r-hat. In this basis, position is beautifully short. The position vector is R times r-hat. All of the direction lives inside that one unit vector. Write those two unit vectors in the fixed frame and the geometry is plain. r-hat leans cos theta along X, plus sine theta along Y. theta-hat swaps the two parts and picks up a minus sign, the fingerprint of a quarter turn.
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Watch them ride the point around the circle. The fixed axes never move. These two turn as the point turns, and that changes everything. That is the price and the power of this frame. The basis depends on where you are. So if we want acceleration, we cannot just differentiate the numbers. We must differentiate the basis vectors too. That is exactly where we go next.
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How fast do the basis vectors themselves change? Differentiate them, chain rule and nothing else. The slide keeps both derivative lines. Notice that every piece carries a theta-dot, because only sweeping turns the basis. Moving straight outward changes neither one. Now look carefully at what those right hand sides really are. They are the unit vectors again, in disguise. The rate of change of r-hat points along theta-hat, scaled by the sweep rate.
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The rate of change of theta-hat points backward, along minus r-hat. Watch the two new arrows appear, each a quarter turn ahead of its parent. Rotation feeds each basis vector into the next. That is the whole machinery. Now differentiate the position itself. The position vector is R times r-hat, so the product rule hands us two pieces. Substitute the basis derivative we just earned, and the velocity lands in its final form on the slide. Clear the scaffolding, and keep the two clean components
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on the point. Read the two pieces as meanings. r-dot is the stretch, how fast the point recedes from the center. R theta-dot is the sweep, the radius times the angular rate. That sweep speed is the omega R from our similar triangles. Every planar velocity is those two, and nothing else.
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One more derivative, and Newton's law learns to turn. Differentiate the velocity term by term, product rule everywhere, and let the slide keep the books. Five raw pieces appear, because every factor takes its own turn. No term dropped, no shortcut taken. Substitute the two basis derivatives we earned, and gather the pieces along r-hat and along theta-hat. Four terms survive. That is the complete acceleration, for every planar motion there has ever been. Four terms, four stories.
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Take them one at a time. The plain pair first. r-double-dot is Newton's law along the radius, nothing more. And R theta-double-dot is its tangential twin, the same law taken along the circle, because R theta measures distance around the rim. Minus R theta-dot squared is the famous one. On a circle it is exactly the V squared over R our similar triangles delivered, the centripetal charge. A rope's tension paid it in lecture two, and a cone wall paid it for a whirling puck.
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Turning always costs something inward. The last term, two r-dot theta-dot, is the stranger. It appears only when the radius changes while the frame sweeps. That is the Coriolis term, the steering hand behind hurricanes and long range artillery. Split any force into radial and tangential parts, and these four terms are the price list. Rotating frames will get an entire lecture of their own.
20:58
A wheel of radius R rolls without slipping, turning at angular rate omega. Mark one point on its rim and follow it. Rolling without slipping means the turn carries the axle forward at exactly R omega, so the center's position is R omega T. Freeze the wheel and read the mark's offsets from the center. The wheel has turned through omega T. So the mark sits R sine omega T behind the center, and R cos omega T below it. The second figure holds that construction still, one small
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right triangle doing all the work. Combine the march and the swing, and the parametric path falls out as a pair. X is the center's advance minus the sine part. Y is the one minus cos height, scaled by R. That curve is the cycloid, a chain of arches, and each arch ends in a sharp cusp that kisses the ground at the point P. Notice the structure. The first term in X is the steady march of the center. The second is the mark swinging forward and back around it.
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Superposition again, the same tool as the projectile, now wrapped around a rolling wheel.
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Differentiate both coordinates to get the velocity. X dot is R omega times, one minus cos omega T. Y dot is R omega sine omega T. Square, add, and use the half angle identity. The speed is two R omega, times the magnitude of sine of half omega T. Read it at the marked points, and the answer is startling. Start at the ground, where the mark touches the road at P. Put in a full turn, and the speed is exactly zero. The bottom of a rolling wheel is momentarily at rest. Now the top.
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The crown does twice the axle's speed. Your tire meets the road at zero, while the tread on top doubles the speedometer. Here is the geometric reason. At every instant the contact point P is the instant center, and the whole wheel is momentarily in pure rotation about it. The center rides a half lever and does R omega. The crown rides the full two R and doubles it. And any point you pick moves perpendicular to its line back to P, faster the longer that line.
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That still point is why tires grip instead of scrub, and why the cusps are perfectly sharp. And take the quick check. Roll the wheel exactly one turn, and the next cusp lands two pi R ahead, the wheel signing its own circumference on the road.
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The final problem, where spinning arrives for real. A bead slides freely on a circular hoop of wire, radius R, no friction anywhere. Spin the whole hoop about its vertical diameter at a steady rate omega, like a coin twirling on a table. Watch it whirl twice. The bead can sit anywhere along the wire. Where can it rest as it spins? The bottom feels obvious. Hold that thought, because the bottom is sometimes wrong. Settle the spin, face the hoop straight on, and choose the frame the motion picks.
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The bead's whole life is one angle, theta, measured from the bottom of the hoop along the wire. As the hoop spins, the bead is dragged around a horizontal circle whose radius is rho, the distance from the spin axis, R sine theta. All the rotation physics lives on that circle. Ride the spinning frame and keep honest books. The wire pushes on the bead with a normal force N, straight along the radius of the hoop. Gravity pulls M G straight down. And the rotation appears as the centrifugal push, M
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omega squared rho, flinging the bead outward from the axis. That push is exactly the centripetal term from the polar box, seen from inside the ride. Now balance along the two directions. The vertical books demand N cos theta match the weight, M G. The horizontal books demand N sine theta supply the fling, M omega squared R sine theta. Divide the second line by the first, and the normal force cancels away entirely. What remains is cos theta equals G over omega squared R.
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One line, and it carries the bead's entire fate.
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A cos can never exceed one. Stare at that. The tilted resting place exists only if G over omega squared R stays at or below one, only if omega squared reaches at least G over R. Spin the hoop too slowly and the equation simply refuses to answer, leaving the bottom as the only resting place. The formula is not broken there. It is telling us the physics changes at a particular spin. That threshold is the critical spin, omega_c, the square root of G over R. Put in a one meter hoop, and omega_c is
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about three radians per second. That is half a turn every second, a lazy twirl you could give it by hand. Below that, the bead sleeps at the bottom. At the threshold, the bottom begins to tremble. Above it, something genuinely new happens. Watch the spin climb. Cross the critical rate and the bottom turns treacherous. What was a valley becomes a summit, and the bead abandons it, climbing the wire to its tilted angle, higher for faster spin. Follow the curve below as it draws itself in step.
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A flat branch along zero, and then, at omega_c, a fork peeling upward. Physicists call that fork a bifurcation, a system quietly rewriting its list of possible fates as one dial turns. And take the quick check at each end. Spin far above critical, and cos theta goes to zero, so the bead rides toward the hoop's equator, where gravity no longer gets a vote. Spin below critical, and the only offer is the bottom. The deepest idea here is stability. Below the critical spin, the bottom is a valley,
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and it holds whatever falls in. Above it, the bottom turns into a hilltop, and the true resting place slides up the wire. Next arc, that idea becomes the entire story. Anything nudged near a stable point oscillates.
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Three habits to carry out of today. First, decompose in the frame the motion picks. When gravity is the only author, use level and plumb, and the stories separate. When things turn, use radius and sweep instead, and the four polar terms sort the physics for you. The right frame turns one hard motion into two solved ones.
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Second, superposition is a weapon, not just a word. Inertia writes a straight story. Gravity writes a falling story. The truth is their sum, added like vectors at every instant, with neither story ever consulting the other. The monkey bet its life against that independence, and lost. Third, geometry answers before calculus. The best angle on a hill came from bisecting two directions. The safe zone came from a vanishing discriminant. Centripetal acceleration came from two similar triangles.
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When the picture speaks first, the calculus becomes a confirmation, not a search.
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Next time, the universal spring. Every valley in physics, once nudged, oscillates, and one equation rules them all. We will find simple harmonic motion hiding inside every potential minimum, including today's tilted valley on the spinning hoop. And we will meet the three fates of damping, under, over, and critical, the difference between a bell, a door closer, and a car's suspension. Bring a steady heartbeat. Next lecture refuses to sit still.
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That wraps up this lecture. This is Amit Sinha. Thanks for watching the Sinha Foundation channel. See you in the next one.
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