{"version":1,"lectureId":"01M14TZXNX6J0FA3SXT86BYP5M","attempt":0,"publication":{"slug":"gyroscopic-precession","title":"Why a Spinning Top Stays Up: Angular Momentum and Precession","subject":"physics","summary":"A vector-first explanation of gyroscopic precession for first-year physics. The lecture replaces the false upward-force intuition with gravitational torque, follows the angular-momentum tip around its circular path, derives the steady-precession rate, tests why faster spin produces slower precession, and closes with a bicycle wheel suspended from one end of its axle.","metaDescription":"See why gravity turns a spinning top sideways, derive its precession rate, and connect the result to a suspended bicycle wheel.","transcript":"Let us begin by admitting that the intuition almost everyone brings to a spinning top is wrong. Spin does not manufacture an upward force, and it does not cancel gravity. Gravity remains present, downward, with the same strength whether the top spins or not. Here is the object we have to explain. One point touches the table at O, the centre of mass C lies above and to one side, and the solid spins rapidly about its tilted axle. This is a three-dimensional arrangement, so let the shape turn once while we identify it. Without spin, the expected motion is simple. The centre of mass drops and the top rotates downward about the contact point. It is tempting to say that spin must provide an opposite, upward effect. But draw the actual force. Gravity acts at the centre of mass and points straight down. The vector r runs from the fixed contact point to that centre. There is still no upward force anywhere in the picture. The distinction between force and torque matters here. Gravity pulls downward, but because its line of action misses O, it also has a turning effect about O. That turning effect is the torque r cross m g. In this pose the torque points sideways, perpendicular to the plane containing r and gravity. That direction is not upward, and it is not the direction in which the centre of mass would fall. Now add the vector created by rapid spin. The angular momentum L points approximately along the axle. The crucial observation is that the gravitational torque is sideways to L, not opposite to it. An opposite torque would reduce the spin angular momentum and bring the top toward rest. A sideways torque does something different: it turns the angular-momentum vector. Since the axle follows that vector in a rapidly spinning symmetric top, the axle turns sideways too. That sideways turning is precession. The top is not held up by a hidden force. It is continually responding to gravity, but its first response is mainly a change of direction around the vertical rather than an immediate collapse toward the table. So our task is now precise. We need to see how a perpendicular torque changes a vector without substantially changing its length, and then calculate how quickly that changing direction travels around the vertical. Forget the physical top for a moment and watch angular momentum in its own vector space. The green arrow has a fixed vertical component and a fixed tilt. Its endpoint is the yellow point. Gravity supplies the magenta torque. At this instant it is tangent to the horizontal circle through the tip of L. In particular, it is perpendicular to L, so it does not point inward or outward along the green arrow. The torque continually changes its direction as the top precesses, but it remains tangent to the same circle. Let the vector move. Its tip traces the blue circle while the magenta arrow rides beside it, always pointing along the next small step. The equation governing this motion is compact: the time derivative of angular momentum equals torque. Torque is not merely associated with a change in L. It is the rate and direction of that change. Over a short time d t, the new angular momentum is the old vector plus a small change d L. The change is torque times d t, so it points in the same tangent direction as the torque. Picture that addition head to tail. The new vector points slightly to one side of the old one. Repeating many tiny additions produces a smooth turn rather than a sequence of visible corners. Because the small change is perpendicular to L, their dot product is zero. Geometrically, adding a little sideways vector turns the long vector. It changes the length only by a second-order amount that disappears in the instantaneous limit. We can state the length result exactly for steady precession. Differentiate L squared. The answer is twice L dot torque, and that dot product vanishes because the two vectors are perpendicular. So the magnitude L stays constant while its direction changes. This is the central correction to the usual intuition. Gravity is not failing to act, and the angular momentum is not frozen. Gravity is steering it. There is also no conflict with the idea that angular momentum resists change. A larger L does not make change impossible. It means that a given sideways change represents a smaller angular turn. Watch another part of the circle. The torque points along the motion of the tip, the tip moves around the vertical, and the length of the green vector remains unchanged throughout. The remaining question is quantitative. A given torque moves the vector tip at a given speed. The larger the circle and the larger L are, the slower the corresponding angular sweep must be. Now return to the top and make the approximation explicit. We consider a nearly symmetric top spinning fast enough that its angular momentum lies approximately along the axle. We also look at steady precession, where the tilt angle remains almost constant. There are two angular rates, and confusing them ruins the argument. Lowercase omega is the rapid spin about the top's own axle. Capital omega is the much slower precession of that axle around the vertical. The centre of mass is a distance ell from the support. Gravity acts downward there, so r cross m g supplies the sideways torque. The angle between the axle and the upward vertical is theta. First calculate the torque magnitude. It is the magnitude of r cross m g. The angle between r and downward gravity is pi minus theta, whose sine is the same as sine theta. Therefore the torque magnitude is m g ell sine theta. That sine factor has a direct geometric meaning. If the axle were vertical, gravity's line of action would pass through the support and the torque would vanish. Tilting the axle gives gravity a perpendicular lever arm ell sine theta. Now calculate how fast the tip of L moves in angular-momentum space. Its horizontal circle has radius L sine theta. A point moving around that circle at angular rate capital omega has speed capital omega times L sine theta. This is ordinary circular-motion geometry, but the circle lives in vector space. The radius is the horizontal component of L, and the speed around it is the magnitude of d L by d t. The dynamical law supplies that vector speed. Since d L by d t equals torque, the speed of the tip must equal the gravitational torque magnitude. Equate them. Capital omega times L sine theta equals m g ell sine theta. This line is the whole balance: the required turning rate of L on the left, and the turning rate supplied by gravity on the right. For a nonzero tilt, sine theta appears on both sides and cancels. The steady-precession rate is therefore m g ell divided by L. The units already make sense. Torque has units of angular momentum per time, so torque divided by L has units of one over time, exactly the units of an angular rate. For rapid spin about a symmetry axis, angular momentum is approximately moment of inertia I times spin rate lowercase omega. A heavier rim gives a larger I, and faster rotation gives a larger omega. Substitute that spin angular momentum. Capital omega is approximately m g ell divided by I times lowercase omega. This is the standard rate for steady precession of a fast symmetric top. The formula has a clean physical reading. More gravitational torque, from more weight or a longer lever arm, turns L faster. More angular momentum, from a larger moment of inertia or faster spin, makes the same torque turn the vector more slowly. Notice what did not survive the cancellation: the tilt angle. In this ideal steady, fast-top model, the leading precession rate is independent of theta. That does not say every real release is steady, or that a top may be tilted arbitrarily without complication. Friction slowly reduces the spin, large wobbling violates the steady approximation, and a badly asymmetric body needs a more complete treatment. Within the model we stated, however, the inverse dependence on spin is a sharp prediction that we can test. The derived formula makes a prediction that initially sounds backward. Spin the top faster, and its slow motion around the vertical becomes slower, not faster. Let the two tops be identical, with the same mass, lever arm, and moment of inertia. Take m g ell divided by I to be twenty per second squared. Only the spin rate changes. The first top spins at one hundred radians per second, giving a precession rate of zero point two radians per second. The second spins twice as fast, at two hundred, and the formula predicts only zero point one. Run both predictions over the same interval. The original top's angular-momentum tip travels one full circuit. The faster-spinning top covers only half a circuit in that time. The pictures are not claiming that the faster top has less angular momentum. It has more. That larger vector is exactly why its direction changes by a smaller angle each second. Gravity has not become weaker, and the torque has not disappeared. For identical geometry it supplies the same sideways change d L in each second. The same change attached to a vector twice as large produces half the angular turn. This is the part that sounds backward only if we use the word rotation for two different things. The top spins rapidly about its own axle, while the axle precesses slowly around the vertical. Increasing the first rate decreases the second. The inverse relationship fits in one line. Capital omega is a fixed top-dependent constant divided by lowercase omega. Doubling lowercase omega doubles the spin angular momentum. Because the gravitational torque is essentially unchanged, it then takes twice as much time to turn L through a given angle. A practical comparison must be made before friction changes the spin very much, and with nearly equal tilt and geometry. Under those conditions, doubling the spin rate should approximately halve the steady-precession rate. As a real top loses spin to friction, the same formula predicts that its precession tends to speed up. Eventually the fast-top approximation fails, the motion becomes more complicated, and the top falls. The apparent stability was never freedom from gravity. It was gravity steering a large angular momentum slowly. A bicycle wheel suspended from one end makes that steering unusually easy to see. The same argument becomes striking with a bicycle wheel. Tie a string to one end of the axle, hold the wheel with its axle horizontal, spin it, and release. The string supplies a fixed support point O while the wheel's centre of mass hangs well to one side. If the wheel is not spinning, there is no mystery. Gravity turns the axle downward about the supported end, and the wheel drops. Now remove that imagined fall and spin the wheel rapidly. Its angular momentum points along the axle. The direction follows the right-hand rule, so reversing the wheel's spin reverses L. Gravity still acts at C. The lever arm r runs from the string's support to the centre, so gravity produces r cross m g. With r horizontal and gravity downward, the torque points sideways. That sideways torque is perpendicular to the wheel's spin angular momentum. It therefore changes the direction of L rather than rapidly reducing its magnitude. The axle follows the turning vector. The unsupported centre does begin to accelerate, but not simply downward. As L turns sideways, the axle turns sideways with it. The continual response to gravity becomes motion around the vertical support line. The wheel therefore swings around the vertical line through O. This is the same precession as the top, now exposed in a form where the single support point and the long lever arm are easy to see. Write the chain once more. The gravitational torque is r cross m g. Torque is d L by d t. For a fast wheel in approximately steady precession, the angular rate is m g ell divided by L. The demonstration also checks direction. Reverse the spin and L reverses, while the gravitational torque for the same pose does not. The sideways change then carries the axle around the vertical in the opposite sense. Spin the wheel faster and L grows. The torque is almost unchanged, so the wheel precesses more slowly. Let friction reduce the spin and the precession generally quickens until the simple steady model ceases to describe the motion well. Real tops usually add a small bobbing of the tilt angle called nutation, caused by how they are released and by departures from perfectly steady precession. We can now answer the opening question without invoking a hidden supporting force. Gravity tries to turn the body about its contact or support point. Rapid spin supplies a large angular-momentum vector, and the gravitational torque changes that vector mainly sideways. The vector tip therefore travels around the vertical, the axle follows it, and we see precession. A top does not refuse gravity. Its spin changes the direction in which gravity first makes the motion develop. That is why a rapidly spinning top stays up for a while, why it circles slowly instead of simply falling, and why making it spin faster slows that circle down.","watch":{"version":1,"scenes":[{"title":"The Intuition That Fails","start":0,"end":155.61575,"objects":{"angular_momentum":"a Vector [green] labelled \"arrow(L)\" drawn in frame (start=(0.0, 0.0, 0.0), end=(2.2, 0.0, 2.64))","axle":"a Line [blue] labelled \"upright(\"spin axis\")\" drawn in frame (start=(0.0, 0.0, 0.0), end=(2.2, 0.0, 2.64))","body":"a Cylinder [blue] drawn in frame (start=(0.45, 0.0, 0.54), end=(1.45, 0.0, 1.74), radius=0.55)","centre":"a Point [yellow] labelled \"C\" drawn in frame (location=(1.0, 0.0, 1.2))","fall_arc":"a CurvedArrow [gray] labelled \"upright(\"fall\")\" drawn in frame (start=(1.45, 0.0, 1.74), end=(0.65, 0.0, 0.45), bend=0.45)","frame":"an Axes3D (x_range=(-3.2, 3.2), y_range=(-3.2, 3.2), z_range=(-3.2, 3.2))","pivot_point":"a Point [text] labelled \"O\" drawn in frame (location=(0.0, 0.0, 0.0))","precession":"a CurvedArrow [yellow] labelled \"Omega\" drawn in frame (start=(1.65, -0.95, 2.05), end=(1.65, 0.95, 2.05), bend=0.75)","question":"a Panel that says \"Why does a spinning top resist falling and instead swing slowly around the vertical?\"","radius":"a Vector [yellow] labelled \"arrow(r)\" drawn in frame (start=(0.0, 0.0, 0.0), end=(1.0, 0.0, 1.2))","right":"a Text [text] that says \"Correct mechanism: gravity changes the direction of angular momentum.\"","torque":"a Vector [magenta] labelled \"arrow(tau)\" drawn in frame (start=(0.0, 0.0, 0.0), end=(0.0, 1.45, 0.0))","vertical":"a Line [gray] labelled \"upright(\"vertical\")\" drawn in frame (start=(0.0, 0.0, 0.0), end=(0.0, 0.0, 2.9), dashed=True)","weight":"a Vector [red] labelled \"m arrow(g)\" drawn in frame (start=(1.0, 0.0, 1.2), end=(1.0, 0.0, -0.55))","wrong":"a Text [text] that says \"Wrong intuition: rapid spin creates an upward force.\""},"beats":[{"start":0,"say":"Let us begin by admitting that the intuition almost everyone brings to a spinning top is wrong. Spin does not manufacture an upward force, and it does not cancel gravity. Gravity remains present, downward, with the same strength whether the top spins or not.","live":[],"does":[[0,"question is shown on the screen, written out."],[16.9045,"question moves to a new place on the board."]]},{"start":17.5045,"say":"Here is the object we have to explain. One point touches the table at O, the centre of mass C lies above and to one side, and the solid spins rapidly about its tilted axle. This is a three-dimensional arrangement, so let the shape turn once while we identify it.","live":["question"],"does":[[17.5045,"frame is shown on the screen, written out."],[18.340000000000003,"body is shown on the screen, written out."],[21.939,"pivot_point is shown on the screen, written out."],[28.186,"axle is shown on the screen, written out."],[29.985000000000003,"vertical is shown on the screen, written out."],[32.33,"frame turns in its own slot."]]},{"start":35.0775,"say":"Without spin, the expected motion is simple. The centre of mass drops and the top rotates downward about the contact point. It is tempting to say that spin must provide an opposite, upward effect.","live":["question","frame","body","axle","pivot_point","vertical"],"does":[[36.73800000000001,"fall_arc is shown on the screen, written out."],[44.551,"frame moves to a new place on the board."],[44.551,"wrong is shown on the screen, written out."]]},{"start":49.11,"say":"But draw the actual force. Gravity acts at the centre of mass and points straight down. The vector r runs from the fixed contact point to that centre. There is still no upward force anywhere in the picture.","live":["question","frame","wrong","body","axle","pivot_point","vertical","fall_arc"],"does":[[49.11,"fall_arc is hidden from the screen."],[52.872,"centre is shown on the screen, written out."],[54.671,"weight is shown on the screen, written out."],[55.995,"radius is shown on the screen, written out."],[61.196,"wrong is indicated — a transient flash."]]},{"start":64.246,"say":"The distinction between force and torque matters here. Gravity pulls downward, but because its line of action misses O, it also has a turning effect about O. That turning effect is the torque r cross m g.","live":["question","frame","wrong","body","axle","pivot_point","vertical","centre","radius","weight"],"does":[[66.12700000000001,"torque is shown on the screen, written out."],[68.042,"weight is indicated — a transient flash."],[76.91300000000001,"radius is indicated — a transient flash."]]},{"start":79.359,"say":"In this pose the torque points sideways, perpendicular to the plane containing r and gravity. That direction is not upward, and it is not the direction in which the centre of mass would fall.","live":["question","frame","wrong","body","axle","pivot_point","vertical","centre","radius","weight","torque"],"does":[[81.611,"torque is indicated — a transient flash."]]},{"start":92.2075,"say":"Now add the vector created by rapid spin. The angular momentum L points approximately along the axle. The crucial observation is that the gravitational torque is sideways to L, not opposite to it.","live":null,"does":[[96.132,"right is shown on the screen, written out."],[97.165,"angular_momentum is shown on the screen, written out."],[103.017,"torque is indicated — a transient flash."],[103.84100000000001,"angular_momentum is indicated — a transient flash."]]},{"start":106.6005,"say":"An opposite torque would reduce the spin angular momentum and bring the top toward rest. A sideways torque does something different: it turns the angular-momentum vector. Since the axle follows that vector in a rapidly spinning symmetric top, the axle turns sideways too.","live":["question","frame","wrong","right","body","axle","pivot_point","vertical","centre","radius","weight","torque","angular_momentum"],"does":[[112.82400000000001,"torque is indicated — a transient flash."],[116.933,"angular_momentum is indicated — a transient flash."]]},{"start":125.3005,"say":"That sideways turning is precession. The top is not held up by a hidden force. It is continually responding to gravity, but its first response is mainly a change of direction around the vertical rather than an immediate collapse toward the table.","live":null,"does":[[127.065,"precession is shown on the screen, written out."],[135.49400000000003,"right is indicated — a transient flash."]]},{"start":141.3415,"say":"So our task is now precise. We need to see how a perpendicular torque changes a vector without substantially changing its length, and then calculate how quickly that changing direction travels around the vertical.","live":["question","frame","wrong","right","body","axle","pivot_point","vertical","centre","radius","weight","torque","angular_momentum","precession"],"does":[[154.57408333333333,"frame is hidden from the screen — left the board."],[154.57408333333333,"body is hidden from the screen — frame left the board."],[154.57408333333333,"axle is hidden from the screen — frame left the board."],[154.57408333333333,"pivot_point is hidden from the screen — frame left the board."],[154.57408333333333,"vertical is hidden from the screen — frame left the board."],[154.57408333333333,"centre is hidden from the screen — frame left the board."],[154.57408333333333,"radius is hidden from the screen — frame left the board."],[154.57408333333333,"weight is hidden from the screen — frame left the board."],[154.57408333333333,"torque is hidden from the screen — frame left the board."],[154.57408333333333,"angular_momentum is hidden from the screen — frame left the board."],[154.57408333333333,"precession is hidden from the screen — frame left the board."],[154.57408333333333,"question is hidden from the screen — left the board."],[154.57408333333333,"right is hidden from the screen — left the board."],[154.57408333333333,"wrong is hidden from the screen — left the board."]]}]},{"title":"The Tip of the Vector","start":155.61575,"end":319.2494583333333,"objects":{"constant":"a Math [text] that says \"$L = abs(arrow(L)) = upright(\"constant\")$\"","heading":"a Heading that says \"A Sideways Change in $arrow(L)$\"","heading_work":"a Heading that says \"Direction Changes, Length Does Not\"","law":"a Math [text] that says \"$frac(dif arrow(L), dif t) = arrow(tau)$\"","magnitude":"a Math [text] that says \"$frac(dif L^2, dif t) = 2 arrow(L) dot arrow(tau) = 0$\"","momentum":"a Vector [green] labelled \"arrow(L)\" drawn in space (start=(0.0, 0.0, 0.0), end=((1.970026884016805 * cos(phase)), (1.970026884016805 * sin(pha…)","phase":"a VariableNumber (initial_value=0.35)","space":"an Axes3D (x_range=(-3.5, 3.5), y_range=(-3.5, 3.5), z_range=(-3.5, 3.5))","tip":"a Point [yellow] drawn in space (location=((1.970026884016805 * cos(phase)), (1.970026884016805 * sin(pha…)","tip_path":"a Circle [blue] labelled \"upright(\"tip path\")\" drawn in space (center=(0.0, 0.0, 1.5391536883141457), radius=1.970026884016805, normal_vector=(0.0, 0.0, 1.0))","torque":"a Vector [magenta] labelled \"arrow(tau)\" drawn in space (start=((1.970026884016805 * cos(phase)), (1.970026884016805 * sin(pha…, end=(((1.970026884016805 * cos(phase)) - (0.85 * sin(phase))), ((1.…)","vector_work":"a Derivation [text] that says \"$arrow(L)(t+dif t) &= arrow(L)(t)+dif arrow(L) \\ dif arrow(L) &= arrow(tau) dif t \\ arrow(L) dot dif arrow(L) &= 0$\"","vertical":"a Line [gray] labelled \"upright(\"vertical\")\" drawn in space (start=(0.0, 0.0, 0.0), end=(0.0, 0.0, 3.0), dashed=True)"},"beats":[{"start":155.61575,"say":"Forget the physical top for a moment and watch angular momentum in its own vector space. The green arrow has a fixed vertical component and a fixed tilt. Its endpoint is the yellow point.","live":[],"does":[[155.61575,"heading is shown on the screen, written out."],[155.61575,"space is shown on the screen, written out."],[161.02575,"momentum is shown on the screen, written out."],[162.34975,"vertical is shown on the screen, written out."],[165.00775,"tip is shown on the screen, written out."]]},{"start":167.41924999999998,"say":"Gravity supplies the magenta torque. At this instant it is tangent to the horizontal circle through the tip of L. In particular, it is perpendicular to L, so it does not point inward or outward along the green arrow.","live":["space","heading","vertical","momentum","tip"],"does":[[169.27675,"torque is shown on the screen, written out."],[176.38174999999998,"momentum is indicated — a transient flash."],[176.38174999999998,"torque is indicated — a transient flash."]]},{"start":181.81175,"say":"The torque continually changes its direction as the top precesses, but it remains tangent to the same circle. Let the vector move. Its tip traces the blue circle while the magenta arrow rides beside it, always pointing along the next small step.","live":["space","heading","vertical","momentum","tip","torque"],"does":[[187.51274999999998,"tip_path is shown on the screen, written out."],[189.50975,"momentum is redrawn as the numbers it depends on change."],[189.50975,"tip is redrawn as the numbers it depends on change."],[189.50975,"torque is redrawn as the numbers it depends on change."],[189.50975,"phase ticks to 6.633185307179586."]]},{"start":198.02724999999998,"say":"The equation governing this motion is compact: the time derivative of angular momentum equals torque. Torque is not merely associated with a change in L. It is the rate and direction of that change.","live":["space","heading","vertical","momentum","tip","torque","tip_path"],"does":[[202.00975,"law is shown on the screen, written out."],[204.22674999999998,"law (the \"arrow(tau)\" part) is emphasized."],[211.35575,"space moves to a new place on the board."],[211.35575,"heading is hidden from the screen — left the board."],[211.35575,"law is hidden from the screen — left the board."],[211.35575,"law (the \"arrow(tau)\" part) is no longer emphasized."]]},{"start":211.95575,"say":"Over a short time d t, the new angular momentum is the old vector plus a small change d L. The change is torque times d t, so it points in the same tangent direction as the torque.","live":["space","vertical","momentum","tip","torque","tip_path"],"does":[[211.95575,"heading_work is shown on the screen, written out."],[214.40575,"vector_work is shown on the screen, written out."],[219.46675,"vector_work is shown on the screen, written out."],[222.11374999999998,"torque is indicated — a transient flash."]]},{"start":224.82725,"say":"Picture that addition head to tail. The new vector points slightly to one side of the old one. Repeating many tiny additions produces a smooth turn rather than a sequence of visible corners.","live":["space","vertical","momentum","tip","torque","tip_path","heading_work"],"does":[[225.91875,"vector_work is indicated — a transient flash."],[232.80374999999998,"momentum is redrawn as the numbers it depends on change."],[232.80374999999998,"tip is redrawn as the numbers it depends on change."],[232.80374999999998,"torque is redrawn as the numbers it depends on change."],[232.80374999999998,"phase ticks to 7.7327427359360135."]]},{"start":236.86925,"say":"Because the small change is perpendicular to L, their dot product is zero. Geometrically, adding a little sideways vector turns the long vector. It changes the length only by a second-order amount that disappears in the instantaneous limit.","live":null,"does":[[239.81275,"vector_work is shown on the screen, written out."],[240.54375,"vector_work (the \"0\" part) is emphasized."],[251.28275,"vector_work (the \"0\" part) is no longer emphasized."]]},{"start":251.88275,"say":"We can state the length result exactly for steady precession. Differentiate L squared. The answer is twice L dot torque, and that dot product vanishes because the two vectors are perpendicular.","live":null,"does":[[255.87675000000002,"magnitude is shown on the screen, written out."],[260.72974999999997,"magnitude (the \"arrow(L) dot arrow(tau)\" part) is emphasized."],[261.30975,"magnitude (the \"0\" part) is emphasized."],[261.30975,"magnitude (the \"arrow(L) dot arrow(tau)\" part) is no longer emphasized."]]},{"start":264.89374999999995,"say":"So the magnitude L stays constant while its direction changes. This is the central correction to the usual intuition. Gravity is not failing to act, and the angular momentum is not frozen. Gravity is steering it.","live":["space","vertical","momentum","tip","torque","tip_path","magnitude","heading_work"],"does":[[264.89374999999995,"magnitude (the \"0\" part) is no longer emphasized."],[265.54375,"constant is shown on the screen, written out."],[278.87175,"A box is drawn around constant."]]},{"start":280.57525,"say":"There is also no conflict with the idea that angular momentum resists change. A larger L does not make change impossible. It means that a given sideways change represents a smaller angular turn.","live":["space","vertical","momentum","tip","torque","tip_path","magnitude","constant","heading_work"],"does":[[286.06674999999996,"momentum is indicated — a transient flash."],[289.95574999999997,"torque is indicated — a transient flash."]]},{"start":293.87725,"say":"Watch another part of the circle. The torque points along the motion of the tip, the tip moves around the vertical, and the length of the green vector remains unchanged throughout.","live":null,"does":[[294.22574999999995,"momentum is redrawn as the numbers it depends on change."],[294.22574999999995,"tip is redrawn as the numbers it depends on change."],[294.22574999999995,"torque is redrawn as the numbers it depends on change."],[294.22574999999995,"phase ticks to 10.874335389525807."],[296.70975,"torque is indicated — a transient flash."],[301.64374999999995,"momentum is indicated — a transient flash."]]},{"start":305.63374999999996,"say":"The remaining question is quantitative. A given torque moves the vector tip at a given speed. The larger the circle and the larger L are, the slower the corresponding angular sweep must be.","live":null,"does":[[318.2077916666666,"constant is hidden from the screen — left the board."],[318.2077916666666,"heading_work is hidden from the screen — left the board."],[318.2077916666666,"magnitude is hidden from the screen — left the board."],[318.2077916666666,"space is hidden from the screen — left the board."],[318.2077916666666,"vertical is hidden from the screen — space left the board."],[318.2077916666666,"momentum is hidden from the screen — space left the board."],[318.2077916666666,"tip is hidden from the screen — space left the board."],[318.2077916666666,"torque is hidden from the screen — space left the board."],[318.2077916666666,"tip_path is hidden from the screen — space left the board."],[318.2077916666666,"vector_work is hidden from the screen — left the board."]]}]},{"title":"Deriving the Precession Rate","start":319.2494583333333,"end":568.5615416666666,"objects":{"axis":"a Line [blue] drawn in frame (start=(0.0, 0.0, 0.0), end=(1.95, 0.0, 2.67))","balance":"a Math [text] that says \"$Omega L sin theta = m g ell sin theta$\"","body":"a Cylinder [blue] drawn in frame (start=(0.4, 0.0, 0.55), end=(1.35, 0.0, 1.85), radius=0.52)","centre":"a Point [yellow] labelled \"C\" drawn in frame (location=(0.95, 0.0, 1.3))","final_rate":"a Math [text] that says \"$Omega approx frac(m g ell, I omega)$\"","frame":"an Axes3D (x_range=(-3.2, 3.2), y_range=(-3.2, 3.2), z_range=(-3.2, 3.2))","heading_balance":"a Heading that says \"Match Torque to the Motion of $arrow(L)$\"","heading_model":"a Heading that says \"The Fast-Top Model\"","heading_rate":"a Heading that says \"The Steady-Precession Rate\"","lever":"a Vector [yellow] labelled \"arrow(r)\" drawn in frame (start=(0.0, 0.0, 0.0), end=(0.95, 0.0, 1.3))","model":"a Panel that says \"For a fast, nearly symmetric top in steady precession, $arrow(L)$ is approximately parallel to the spin axis and the tilt angle is nearly constant.\"","momentum":"a Vector [green] labelled \"arrow(L)\" drawn in frame (start=(0.0, 0.0, 0.0), end=(2.25, 0.0, 3.08))","notation":"a Math [text] that says \"$omega = upright(\"spin\"), quad Omega = upright(\"precession\")$\"","pivot_point":"a Point [text] labelled \"O\" drawn in frame (location=(0.0, 0.0, 0.0))","precession_work":"a Derivation [text] that says \"$abs(frac(dif arrow(L), dif t)) &= Omega (L sin theta) \\ &= abs(arrow(tau))$\"","reduced":"a Math [text] that says \"$Omega = frac(m g ell, L)$\"","spin_momentum":"a Math [text] that says \"$L approx I omega$\"","tilt":"an Angle [yellow] labelled \"theta\" drawn in frame (vertex=(0.0, 0.0, 0.0), sides=((1.95, 0.0, 2.67), (0.0, 0.0, 3.0)))","tip_circle":"a Circle [cyan] labelled \"upright(\"path of tip\")\" drawn in frame (center=(0.0, 0.0, 3.08), radius=2.25, normal_vector=(0.0, 0.0, 1.0))","torque":"a Vector [magenta] labelled \"arrow(tau)\" drawn in frame (start=(0.0, 0.0, 0.0), end=(0.0, 1.35, 0.0))","torque_work":"a Derivation [text] that says \"$abs(arrow(tau)) &= abs(arrow(r) times m arrow(g)) \\ &= m g ell sin theta$\"","vertical":"a Line [gray] labelled \"upright(\"vertical\")\" drawn in frame (start=(0.0, 0.0, 0.0), end=(0.0, 0.0, 3.0), dashed=True)","weight":"a Vector [red] labelled \"m arrow(g)\" drawn in frame (start=(0.95, 0.0, 1.3), end=(0.95, 0.0, -0.55))"},"beats":[{"start":319.2494583333333,"say":"Now return to the top and make the approximation explicit. We consider a nearly symmetric top spinning fast enough that its angular momentum lies approximately along the axle. We also look at steady precession, where the tilt angle remains almost constant.","live":[],"does":[[319.2494583333333,"heading_model is shown on the screen, written out."],[319.2494583333333,"frame is shown on the screen, written out."],[320.1434583333333,"body is shown on the screen, written out."],[320.1434583333333,"pivot_point is shown on the screen, written out."],[324.4504583333333,"centre is shown on the screen, written out."],[327.2954583333333,"momentum is shown on the screen, written out."],[329.77945833333325,"axis is shown on the screen, written out."],[331.8814583333333,"frame moves to a new place on the board."],[331.8814583333333,"model is shown on the screen, written out."],[333.4604583333333,"vertical is shown on the screen, written out."],[333.4604583333333,"tilt is shown on the screen, written out."]]},{"start":336.6144583333333,"say":"There are two angular rates, and confusing them ruins the argument. Lowercase omega is the rapid spin about the top's own axle. Capital omega is the much slower precession of that axle around the vertical.","live":["model","frame","heading_model","body","axis","vertical","pivot_point","centre","momentum","tilt"],"does":[[337.4384583333333,"notation is shown on the screen, written out."],[341.6414583333333,"notation (the \"omega\" part) is emphasized."],[346.4244583333333,"notation (the \"Omega\" part) is emphasized."],[346.4244583333333,"notation (the \"omega\" part) is no longer emphasized."],[351.4629583333333,"notation (the \"Omega\" part) is no longer emphasized."]]},{"start":352.0629583333333,"say":"The centre of mass is a distance ell from the support. Gravity acts downward there, so r cross m g supplies the sideways torque. The angle between the axle and the upward vertical is theta.","live":["model","notation","frame","heading_model","body","axis","vertical","pivot_point","centre","momentum","tilt"],"does":[[353.5724583333333,"lever is shown on the screen, written out."],[355.8714583333333,"weight is shown on the screen, written out."],[360.6194583333333,"torque is shown on the screen, written out."],[364.6944583333333,"tilt is indicated — a transient flash."],[365.5424583333333,"frame moves to a new place on the board."],[365.5424583333333,"heading_model is hidden from the screen — left the board."],[365.5424583333333,"model is hidden from the screen — left the board."],[365.5424583333333,"notation is hidden from the screen — left the board."]]},{"start":366.14245833333325,"say":"First calculate the torque magnitude. It is the magnitude of r cross m g. The angle between r and downward gravity is pi minus theta, whose sine is the same as sine theta. Therefore the torque magnitude is m g ell sine theta.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque"],"does":[[366.14245833333325,"heading_balance is shown on the screen, written out."],[367.5234583333333,"torque_work is shown on the screen, written out."],[370.64645833333327,"lever is indicated — a transient flash."],[374.7684583333333,"weight is indicated — a transient flash."],[379.8424583333333,"torque_work is shown on the screen, written out."]]},{"start":384.3314583333333,"say":"That sine factor has a direct geometric meaning. If the axle were vertical, gravity's line of action would pass through the support and the torque would vanish. Tilting the axle gives gravity a perpendicular lever arm ell sine theta.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque","heading_balance"],"does":[[394.95445833333326,"tilt is indicated — a transient flash."],[397.60145833333326,"torque_work (the \"ell sin theta\" part) is emphasized."],[399.9579583333333,"torque_work (the \"ell sin theta\" part) is no longer emphasized."]]},{"start":400.55795833333326,"say":"Now calculate how fast the tip of L moves in angular-momentum space. Its horizontal circle has radius L sine theta. A point moving around that circle at angular rate capital omega has speed capital omega times L sine theta.","live":null,"does":[[401.1844583333333,"precession_work is shown on the screen, written out."],[406.96645833333326,"tip_circle is shown on the screen, written out."],[407.66345833333327,"precession_work (the \"L sin theta\" part) is emphasized."],[412.0164583333333,"precession_work (the \"L sin theta\" part) is no longer emphasized."],[412.0164583333333,"precession_work (the \"Omega\" part) is emphasized."],[417.65895833333326,"precession_work (the \"Omega\" part) is no longer emphasized."]]},{"start":418.2589583333333,"say":"This is ordinary circular-motion geometry, but the circle lives in vector space. The radius is the horizontal component of L, and the speed around it is the magnitude of d L by d t.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque","heading_balance","tip_circle"],"does":[[421.9394583333333,"tip_circle is indicated — a transient flash."],[426.5944583333333,"momentum is indicated — a transient flash."]]},{"start":431.8159583333333,"say":"The dynamical law supplies that vector speed. Since d L by d t equals torque, the speed of the tip must equal the gravitational torque magnitude.","live":null,"does":[[437.5394583333333,"precession_work is shown on the screen, written out."],[439.4554583333333,"torque is indicated — a transient flash."]]},{"start":442.7254583333333,"say":"Equate them. Capital omega times L sine theta equals m g ell sine theta. This line is the whole balance: the required turning rate of L on the left, and the turning rate supplied by gravity on the right.","live":null,"does":[[443.0734583333333,"balance is shown on the screen, written out."],[454.0104583333333,"balance (the \"Omega L sin theta\" part) is emphasized."],[457.4584583333333,"balance (the \"Omega L sin theta\" part) is no longer emphasized."],[457.4584583333333,"balance (the \"m g ell sin theta\" part) is emphasized."],[459.59495833333324,"balance moves to a new place on the board."],[459.59495833333324,"heading_balance is hidden from the screen — left the board."],[459.59495833333324,"precession_work is hidden from the screen — left the board."],[459.59495833333324,"torque_work is hidden from the screen — left the board."],[459.59495833333324,"balance (the \"m g ell sin theta\" part) is no longer emphasized."]]},{"start":460.19495833333326,"say":"For a nonzero tilt, sine theta appears on both sides and cancels. The steady-precession rate is therefore m g ell divided by L.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque","balance","tip_circle"],"does":[[460.19495833333326,"heading_rate is shown on the screen, written out."],[464.6874583333333,"balance (the \"sin theta\" part) is slashed through — it cancels."],[464.6874583333333,"balance (the \"sin theta#2\" part) is slashed through — it cancels."],[467.35845833333326,"reduced is shown on the screen, written out."],[469.8424583333333,"reduced (the \"L\" part) is emphasized."]]},{"start":471.1859583333333,"say":"The units already make sense. Torque has units of angular momentum per time, so torque divided by L has units of one over time, exactly the units of an angular rate.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque","balance","tip_circle","reduced","heading_rate"],"does":[[471.1859583333333,"reduced (the \"L\" part) is no longer emphasized."],[482.0634583333333,"reduced is indicated — a transient flash."]]},{"start":483.8944583333333,"say":"For rapid spin about a symmetry axis, angular momentum is approximately moment of inertia I times spin rate lowercase omega. A heavier rim gives a larger I, and faster rotation gives a larger omega.","live":null,"does":[[487.0404583333333,"spin_momentum is shown on the screen, written out."],[490.0364583333333,"spin_momentum (the \"I\" part) is emphasized."],[492.19545833333336,"spin_momentum (the \"I\" part) is no longer emphasized."],[492.19545833333336,"spin_momentum (the \"omega\" part) is emphasized."],[498.6969583333333,"spin_momentum (the \"omega\" part) is no longer emphasized."]]},{"start":499.2969583333333,"say":"Substitute that spin angular momentum. Capital omega is approximately m g ell divided by I times lowercase omega. This is the standard rate for steady precession of a fast symmetric top.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque","balance","tip_circle","reduced","spin_momentum","heading_rate"],"does":[[499.6454583333333,"final_rate is shown on the screen, written out."],[509.93145833333335,"A box is drawn around final_rate."]]},{"start":514.3864583333333,"say":"The formula has a clean physical reading. More gravitational torque, from more weight or a longer lever arm, turns L faster. More angular momentum, from a larger moment of inertia or faster spin, makes the same torque turn the vector more slowly.","live":["frame","body","axis","vertical","pivot_point","centre","momentum","tilt","lever","weight","torque","balance","tip_circle","reduced","spin_momentum","final_rate","heading_rate"],"does":[[518.8334583333333,"final_rate (the \"m g ell\" part) is emphasized."],[524.8814583333333,"final_rate (the \"I omega\" part) is emphasized."],[524.8814583333333,"final_rate (the \"m g ell\" part) is no longer emphasized."],[532.4634583333333,"final_rate (the \"I omega\" part) is no longer emphasized."]]},{"start":533.0634583333333,"say":"Notice what did not survive the cancellation: the tilt angle. In this ideal steady, fast-top model, the leading precession rate is independent of theta. That does not say every real release is steady, or that a top may be tilted arbitrarily without complication.","live":null,"does":[[536.3374583333333,"tilt is indicated — a transient flash."]]},{"start":551.0789583333333,"say":"Friction slowly reduces the spin, large wobbling violates the steady approximation, and a badly asymmetric body needs a more complete treatment. Within the model we stated, however, the inverse dependence on spin is a sharp prediction that we can test.","live":null,"does":[[567.519875,"balance is hidden from the screen — left the board."],[567.519875,"final_rate is hidden from the screen — left the board."],[567.519875,"frame is hidden from the screen — left the board."],[567.519875,"body is hidden from the screen — frame left the board."],[567.519875,"axis is hidden from the screen — frame left the board."],[567.519875,"vertical is hidden from the screen — frame left the board."],[567.519875,"pivot_point is hidden from the screen — frame left the board."],[567.519875,"centre is hidden from the screen — frame left the board."],[567.519875,"momentum is hidden from the screen — frame left the board."],[567.519875,"tilt is hidden from the screen — frame left the board."],[567.519875,"lever is hidden from the screen — frame left the board."],[567.519875,"weight is hidden from the screen — frame left the board."],[567.519875,"torque is hidden from the screen — frame left the board."],[567.519875,"tip_circle is hidden from the screen — frame left the board."],[567.519875,"heading_rate is hidden from the screen — left the board."],[567.519875,"reduced is hidden from the screen — left the board."],[567.519875,"spin_momentum is hidden from the screen — left the board."]]}]},{"title":"Faster Spin, Slower Precession","start":568.5615416666666,"end":729.0074999999999,"objects":{"explanation":"a Panel that says \"Doubling the spin doubles $L$. Gravity supplies essentially the same torque, so it takes twice as long to turn the angular-momentum vector through the same angle.\"","fast_caption":"a Tex [text] that says \"Twice the spin\"","fast_centre":"a Point [gray] drawn in fast_figure","fast_circle":"a Circle [blue] drawn in fast_figure","fast_figure":"a Figure (x_range=(-1.4, 1.4), y_range=(-1.4, 1.4), aspect=(1.0, 1.0))","fast_phase":"a VariableNumber (initial_value=0.2)","fast_radius":"a Line [green] drawn in fast_figure (end=((0.0 + (1.0 * cos((((fast_phase * 180.0) / 3.141592653589793) …)","fast_tip":"a Point [yellow] labelled \"arrow(L)\" drawn in fast_figure (location=((0.0 + (1.0 * cos((((fast_phase * 180.0) / 3.141592653589793) …)","fast_value":"a Math [text] that says \"$omega=200 quad => quad Omega=0.10$\"","heading":"a Heading that says \"One Prediction That Sounds Backward\"","heading_result":"a Heading that says \"Why Faster Means Slower\"","inverse":"a Math [text] that says \"$Omega = frac(upright(\"constant\"), omega)$\"","slow_caption":"a Tex [text] that says \"Original spin\"","slow_centre":"a Point [gray] drawn in slow_figure","slow_circle":"a Circle [blue] drawn in slow_figure","slow_figure":"a Figure (x_range=(-1.4, 1.4), y_range=(-1.4, 1.4), aspect=(1.0, 1.0))","slow_phase":"a VariableNumber (initial_value=0.2)","slow_radius":"a Line [green] drawn in slow_figure (end=((0.0 + (1.0 * cos((((slow_phase * 180.0) / 3.141592653589793) …)","slow_tip":"a Point [yellow] labelled \"arrow(L)\" drawn in slow_figure (location=((0.0 + (1.0 * cos((((slow_phase * 180.0) / 3.141592653589793) …)","slow_value":"a Math [text] that says \"$omega=100 quad => quad Omega=0.20$\""},"beats":[{"start":568.5615416666666,"say":"The derived formula makes a prediction that initially sounds backward. Spin the top faster, and its slow motion around the vertical becomes slower, not faster.","live":[],"does":[[568.5615416666666,"heading is shown on the screen, written out."],[573.5185416666666,"slow_caption is shown on the screen, written out."],[573.5185416666666,"fast_caption is shown on the screen, written out."]]},{"start":579.0765416666666,"say":"Let the two tops be identical, with the same mass, lever arm, and moment of inertia. Take m g ell divided by I to be twenty per second squared. Only the spin rate changes.","live":["slow_caption","fast_caption","heading"],"does":[[579.0765416666666,"slow_figure is shown on the screen, written out."],[579.0765416666666,"fast_figure is shown on the screen, written out."],[580.5505416666666,"slow_circle is shown on the screen, written out."],[580.5505416666666,"fast_circle is shown on the screen, written out."],[581.7235416666666,"slow_centre is shown on the screen, written out."],[581.7235416666666,"fast_centre is shown on the screen, written out."]]},{"start":592.2850416666665,"say":"The first top spins at one hundred radians per second, giving a precession rate of zero point two radians per second. The second spins twice as fast, at two hundred, and the formula predicts only zero point one.","live":["slow_figure","slow_caption","fast_figure","fast_caption","heading","slow_circle","fast_circle","slow_centre","fast_centre"],"does":[[592.7955416666666,"slow_tip is shown on the screen, written out."],[592.7955416666666,"slow_radius is shown on the screen, written out."],[595.0835416666666,"fast_tip is shown on the screen, written out."],[595.0835416666666,"fast_radius is shown on the screen, written out."],[597.2305416666666,"slow_value is shown on the screen, written out."],[604.7425416666666,"fast_value is shown on the screen, written out."]]},{"start":606.7820416666666,"say":"Run both predictions over the same interval. The original top's angular-momentum tip travels one full circuit. The faster-spinning top covers only half a circuit in that time.","live":["slow_figure","slow_value","slow_caption","fast_figure","fast_value","fast_caption","heading","slow_circle","fast_circle","slow_centre","fast_centre","slow_tip","slow_radius","fast_tip","fast_radius"],"does":[[608.6285416666666,"slow_tip is redrawn as the numbers it depends on change."],[608.6285416666666,"slow_radius is redrawn as the numbers it depends on change."],[608.6285416666666,"fast_tip is redrawn as the numbers it depends on change."],[608.6285416666666,"fast_radius is redrawn as the numbers it depends on change."],[608.6285416666666,"slow_phase ticks to 6.483185307179586."],[608.6285416666666,"fast_phase ticks to 3.3415926535897933."]]},{"start":619.2935416666666,"say":"The pictures are not claiming that the faster top has less angular momentum. It has more. That larger vector is exactly why its direction changes by a smaller angle each second.","live":null,"does":[[626.4685416666666,"fast_radius is indicated — a transient flash."],[629.4525416666665,"fast_tip is indicated — a transient flash."]]},{"start":632.1075416666666,"say":"Gravity has not become weaker, and the torque has not disappeared. For identical geometry it supplies the same sideways change d L in each second. The same change attached to a vector twice as large produces half the angular turn.","live":null,"does":[]},{"start":648.7060416666666,"say":"This is the part that sounds backward only if we use the word rotation for two different things. The top spins rapidly about its own axle, while the axle precesses slowly around the vertical. Increasing the first rate decreases the second.","live":null,"does":[[663.6830416666666,"fast_caption is hidden from the screen — left the board."],[663.6830416666666,"fast_figure is hidden from the screen — left the board."],[663.6830416666666,"fast_circle is hidden from the screen — fast_figure left the board."],[663.6830416666666,"fast_centre is hidden from the screen — fast_figure left the board."],[663.6830416666666,"fast_tip is hidden from the screen — fast_figure left the board."],[663.6830416666666,"fast_radius is hidden from the screen — fast_figure left the board."],[663.6830416666666,"fast_value is hidden from the screen — left the board."],[663.6830416666666,"heading is hidden from the screen — left the board."],[663.6830416666666,"slow_caption is hidden from the screen — left the board."],[663.6830416666666,"slow_figure is hidden from the screen — left the board."],[663.6830416666666,"slow_circle is hidden from the screen — slow_figure left the board."],[663.6830416666666,"slow_centre is hidden from the screen — slow_figure left the board."],[663.6830416666666,"slow_tip is hidden from the screen — slow_figure left the board."],[663.6830416666666,"slow_radius is hidden from the screen — slow_figure left the board."],[663.6830416666666,"slow_value is hidden from the screen — left the board."]]},{"start":664.8830416666666,"say":"The inverse relationship fits in one line. Capital omega is a fixed top-dependent constant divided by lowercase omega.","live":[],"does":[[664.8830416666666,"heading_result is shown on the screen, written out."],[665.4875416666665,"inverse is shown on the screen, written out."],[672.2785416666666,"inverse (the \"omega\" part) is emphasized."]]},{"start":674.2720416666666,"say":"Doubling lowercase omega doubles the spin angular momentum. Because the gravitational torque is essentially unchanged, it then takes twice as much time to turn L through a given angle.","live":["inverse","heading_result"],"does":[[674.2720416666666,"inverse (the \"omega\" part) is no longer emphasized."],[674.6205416666666,"explanation is shown on the screen, written out."],[683.2225416666665,"inverse is indicated — a transient flash."]]},{"start":686.8185416666665,"say":"A practical comparison must be made before friction changes the spin very much, and with nearly equal tilt and geometry. Under those conditions, doubling the spin rate should approximately halve the steady-precession rate.","live":["inverse","explanation","heading_result"],"does":[[698.9035416666666,"inverse is indicated — a transient flash."]]},{"start":701.3965416666665,"say":"As a real top loses spin to friction, the same formula predicts that its precession tends to speed up. Eventually the fast-top approximation fails, the motion becomes more complicated, and the top falls.","live":null,"does":[]},{"start":715.2785416666666,"say":"The apparent stability was never freedom from gravity. It was gravity steering a large angular momentum slowly. A bicycle wheel suspended from one end makes that steering unusually easy to see.","live":null,"does":[[727.9658333333332,"explanation is hidden from the screen — left the board."],[727.9658333333332,"heading_result is hidden from the screen — left the board."],[727.9658333333332,"inverse is hidden from the screen — left the board."]]}]},{"title":"The Bicycle Wheel on a String","start":729.0074999999999,"end":923.5685624999999,"objects":{"axle":"a Line [blue] labelled \"upright(\"axle\")\" drawn in frame (start=(0.0, 0.0, 1.6), end=(2.65, 0.0, 1.6))","centre":"a Point [text] labelled \"C\" drawn in frame (location=(2.0, 0.0, 1.6))","fall":"a CurvedArrow [gray] labelled \"upright(\"fall\")\" drawn in frame (start=(2.05, 0.0, 1.55), end=(1.15, 0.0, 0.45))","frame":"an Axes3D (x_range=(-1.0, 5.0), y_range=(-3.0, 3.0), z_range=(-1.0, 5.0))","gravity":"a Vector [red] labelled \"m arrow(g)\" drawn in frame (start=(2.0, 0.0, 1.6), end=(2.0, 0.0, 0.1))","heading":"a Heading that says \"A Wheel Suspended from One End\"","heading_work":"a Heading that says \"The Same Vector Argument\"","lever":"a Vector [yellow] labelled \"arrow(r)\" drawn in frame (start=(0.0, 0.0, 1.6), end=(2.0, 0.0, 1.6))","momentum":"a Vector [green] labelled \"arrow(L)\" drawn in frame (start=(2.0, 0.0, 1.6), end=(3.65, 0.0, 1.6))","nutation_note":"a Text [text] that says \"Nutation: a small bobbing of the tilt angle.\"","precession":"a CurvedArrow [yellow] labelled \"Omega\" drawn in frame (start=(1.75, -1.25, 1.6), end=(1.75, 1.25, 1.6), bend=0.8)","result":"a Math [text] that says \"$arrow(tau) arrow.r dif arrow(L) arrow.r upright(\"precession\")$\"","string":"a Line [gray] labelled \"upright(\"string\")\" drawn in frame (start=(0.0, 0.0, 4.2), end=(0.0, 0.0, 1.6))","support_point":"a Point [yellow] labelled \"O\" drawn in frame (location=(0.0, 0.0, 1.6))","torque":"a Vector [magenta] labelled \"arrow(tau)\" drawn in frame (start=(0.0, 0.0, 1.6), end=(0.0, 1.35, 1.6))","wheel":"a Cylinder [blue] drawn in frame (start=(1.82, 0.0, 1.6), end=(2.18, 0.0, 1.6), opacity=0.22)","work":"a Derivation [text] that says \"$arrow(tau) &= arrow(r) times m arrow(g) \\ frac(dif arrow(L), dif t) &= arrow(tau) \\ Omega &approx frac(m g ell, L)$\""},"beats":[{"start":729.0074999999999,"say":"The same argument becomes striking with a bicycle wheel. Tie a string to one end of the axle, hold the wheel with its axle horizontal, spin it, and release. The string supplies a fixed support point O while the wheel's centre of mass hangs well to one side.","live":[],"does":[[729.0074999999999,"heading is shown on the screen, written out."],[729.0074999999999,"frame is shown on the screen, written out."],[731.3294999999999,"frame turns in its own slot."],[731.3294999999999,"wheel is shown on the screen, written out."],[732.5835,"string is shown on the screen, written out."],[733.6864999999999,"axle is shown on the screen, written out."],[741.3024999999999,"support_point is shown on the screen, written out."],[742.6724999999999,"centre is shown on the screen, written out."]]},{"start":745.896,"say":"If the wheel is not spinning, there is no mystery. Gravity turns the axle downward about the supported end, and the wheel drops.","live":["frame","heading","string","support_point","axle","wheel","centre"],"does":[[746.9064999999999,"fall is shown on the screen, written out."]]},{"start":754.2515,"say":"Now remove that imagined fall and spin the wheel rapidly. Its angular momentum points along the axle. The direction follows the right-hand rule, so reversing the wheel's spin reverses L.","live":["frame","heading","string","support_point","axle","wheel","centre","fall"],"does":[[754.9594999999999,"fall is hidden from the screen."],[759.0464999999999,"momentum is shown on the screen, written out."],[766.1284999999999,"momentum is indicated — a transient flash."]]},{"start":767.483,"say":"Gravity still acts at C. The lever arm r runs from the string's support to the centre, so gravity produces r cross m g. With r horizontal and gravity downward, the torque points sideways.","live":["frame","heading","string","support_point","axle","wheel","centre","momentum"],"does":[[767.8314999999999,"gravity is shown on the screen, written out."],[767.8314999999999,"gravity is indicated — a transient flash."],[770.3274999999999,"lever is shown on the screen, written out."],[770.9544999999999,"lever is indicated — a transient flash."],[779.7665,"torque is shown on the screen, written out."]]},{"start":782.0034999999999,"say":"That sideways torque is perpendicular to the wheel's spin angular momentum. It therefore changes the direction of L rather than rapidly reducing its magnitude. The axle follows the turning vector.","live":["frame","heading","string","support_point","axle","wheel","centre","momentum","lever","gravity","torque"],"does":[[783.6054999999999,"torque is indicated — a transient flash."],[788.5055,"momentum is indicated — a transient flash."]]},{"start":795.6004999999999,"say":"The unsupported centre does begin to accelerate, but not simply downward. As L turns sideways, the axle turns sideways with it. The continual response to gravity becomes motion around the vertical support line.","live":null,"does":[[807.5174999999999,"precession is shown on the screen, written out."],[808.3884999999999,"support_point is indicated — a transient flash."]]},{"start":810.0794999999999,"say":"The wheel therefore swings around the vertical line through O. This is the same precession as the top, now exposed in a form where the single support point and the long lever arm are easy to see.","live":["frame","heading","string","support_point","axle","wheel","centre","momentum","lever","gravity","torque","precession"],"does":[[815.0954999999999,"precession is indicated — a transient flash."],[820.0065,"lever is indicated — a transient flash."],[822.0964999999999,"frame moves to a new place on the board."],[822.0964999999999,"heading is hidden from the screen — left the board."]]},{"start":822.6964999999999,"say":"Write the chain once more. The gravitational torque is r cross m g. Torque is d L by d t. For a fast wheel in approximately steady precession, the angular rate is m g ell divided by L.","live":["frame","string","support_point","axle","wheel","centre","momentum","lever","gravity","torque","precession"],"does":[[822.6964999999999,"heading_work is shown on the screen, written out."],[826.0864999999999,"work is shown on the screen, written out."],[826.0864999999999,"work is shown on the screen, written out."],[834.9685,"work is shown on the screen, written out."]]},{"start":838.703,"say":"The demonstration also checks direction. Reverse the spin and L reverses, while the gravitational torque for the same pose does not. The sideways change then carries the axle around the vertical in the opposite sense.","live":["frame","string","support_point","axle","wheel","centre","momentum","lever","gravity","torque","precession","heading_work"],"does":[[843.0455,"momentum is indicated — a transient flash."],[846.7375,"torque is indicated — a transient flash."],[851.5664999999999,"precession is indicated — a transient flash."]]},{"start":853.583,"say":"Spin the wheel faster and L grows. The torque is almost unchanged, so the wheel precesses more slowly. Let friction reduce the spin and the precession generally quickens until the simple steady model ceases to describe the motion well.","live":null,"does":[[855.6264999999999,"work (the \"L\" part) is emphasized."],[869.257,"work (the \"L\" part) is no longer emphasized."]]},{"start":869.857,"say":"Real tops usually add a small bobbing of the tilt angle called nutation, caused by how they are released and by departures from perfectly steady precession.","live":null,"does":[[874.0484999999999,"nutation_note is shown on the screen, written out."]]},{"start":880.3604999999999,"say":"We can now answer the opening question without invoking a hidden supporting force. Gravity tries to turn the body about its contact or support point. Rapid spin supplies a large angular-momentum vector, and the gravitational torque changes that vector mainly sideways.","live":["frame","string","support_point","axle","wheel","centre","momentum","lever","gravity","torque","precession","nutation_note","heading_work"],"does":[[885.7824999999999,"result is shown on the screen, written out."],[885.7824999999999,"work is indicated — a transient flash."],[896.5674999999999,"work is indicated — a transient flash."]]},{"start":898.2244999999999,"say":"The vector tip therefore travels around the vertical, the axle follows it, and we see precession. A top does not refuse gravity. Its spin changes the direction in which gravity first makes the motion develop.","live":["frame","string","support_point","axle","wheel","centre","momentum","lever","gravity","torque","precession","result","nutation_note","heading_work"],"does":[[903.4364999999999,"result is indicated — a transient flash."],[906.3854999999999,"A box is drawn around result."]]},{"start":912.5704999999999,"say":"That is why a rapidly spinning top stays up for a while, why it circles slowly instead of simply falling, and why making it spin faster slows that circle down.","live":null,"does":[[922.5268958333332,"frame is hidden from the screen — left the board."],[922.5268958333332,"string is hidden from the screen — frame left the board."],[922.5268958333332,"support_point is hidden from the screen — frame left the board."],[922.5268958333332,"axle is hidden from the screen — frame left the board."],[922.5268958333332,"wheel is hidden from the screen — frame left the board."],[922.5268958333332,"centre is hidden from the screen — frame left the board."],[922.5268958333332,"momentum is hidden from the screen — frame left the board."],[922.5268958333332,"lever is hidden from the screen — frame left the board."],[922.5268958333332,"gravity is hidden from the screen — frame left the board."],[922.5268958333332,"torque is hidden from the screen — frame left the board."],[922.5268958333332,"precession is hidden from the screen — frame left the board."],[922.5268958333332,"heading_work is hidden from the screen — left the board."],[922.5268958333332,"nutation_note is hidden from the screen — left the board."],[922.5268958333332,"result is hidden from the screen — left the board."],[922.5268958333332,"work is hidden from the screen — left the board."]]}]}]},"durationSeconds":924,"chapters":[{"title":"The Intuition That Fails","startSeconds":0,"narration":"Let us begin by admitting that the intuition almost everyone brings to a spinning top is wrong. Spin does not manufacture an upward force, and it does not cancel gravity. Gravity remains present, downward, with the same strength whether the top spins or not. Here is the object we have to explain. One point touches the table at O, the centre of mass C lies above and to one side, and the solid spins rapidly about its tilted axle. This is a three-dimensional arrangement, so let the shape turn once while we identify it. Without spin, the expected motion is simple. The centre of mass drops and the top rotates downward about the contact point. It is tempting to say that spin must provide an opposite, upward effect. But draw the actual force. Gravity acts at the centre of mass and points straight down. The vector r runs from the fixed contact point to that centre. There is still no upward force anywhere in the picture. The distinction between force and torque matters here. Gravity pulls downward, but because its line of action misses O, it also has a turning effect about O. That turning effect is the torque r cross m g. In this pose the torque points sideways, perpendicular to the plane containing r and gravity. That direction is not upward, and it is not the direction in which the centre of mass would fall. Now add the vector created by rapid spin. The angular momentum L points approximately along the axle. The crucial observation is that the gravitational torque is sideways to L, not opposite to it. An opposite torque would reduce the spin angular momentum and bring the top toward rest. A sideways torque does something different: it turns the angular-momentum vector. Since the axle follows that vector in a rapidly spinning symmetric top, the axle turns sideways too. That sideways turning is precession. The top is not held up by a hidden force. It is continually responding to gravity, but its first response is mainly a change of direction around the vertical rather than an immediate collapse toward the table. So our task is now precise. We need to see how a perpendicular torque changes a vector without substantially changing its length, and then calculate how quickly that changing direction travels around the vertical."},{"title":"The Tip of the Vector","startSeconds":155.61575,"narration":"Forget the physical top for a moment and watch angular momentum in its own vector space. The green arrow has a fixed vertical component and a fixed tilt. Its endpoint is the yellow point. Gravity supplies the magenta torque. At this instant it is tangent to the horizontal circle through the tip of L. In particular, it is perpendicular to L, so it does not point inward or outward along the green arrow. The torque continually changes its direction as the top precesses, but it remains tangent to the same circle. Let the vector move. Its tip traces the blue circle while the magenta arrow rides beside it, always pointing along the next small step. The equation governing this motion is compact: the time derivative of angular momentum equals torque. Torque is not merely associated with a change in L. It is the rate and direction of that change. Over a short time d t, the new angular momentum is the old vector plus a small change d L. The change is torque times d t, so it points in the same tangent direction as the torque. Picture that addition head to tail. The new vector points slightly to one side of the old one. Repeating many tiny additions produces a smooth turn rather than a sequence of visible corners. Because the small change is perpendicular to L, their dot product is zero. Geometrically, adding a little sideways vector turns the long vector. It changes the length only by a second-order amount that disappears in the instantaneous limit. We can state the length result exactly for steady precession. Differentiate L squared. The answer is twice L dot torque, and that dot product vanishes because the two vectors are perpendicular. So the magnitude L stays constant while its direction changes. This is the central correction to the usual intuition. Gravity is not failing to act, and the angular momentum is not frozen. Gravity is steering it. There is also no conflict with the idea that angular momentum resists change. A larger L does not make change impossible. It means that a given sideways change represents a smaller angular turn. Watch another part of the circle. The torque points along the motion of the tip, the tip moves around the vertical, and the length of the green vector remains unchanged throughout. The remaining question is quantitative. A given torque moves the vector tip at a given speed. The larger the circle and the larger L are, the slower the corresponding angular sweep must be."},{"title":"Deriving the Precession Rate","startSeconds":319.2494583333333,"narration":"Now return to the top and make the approximation explicit. We consider a nearly symmetric top spinning fast enough that its angular momentum lies approximately along the axle. We also look at steady precession, where the tilt angle remains almost constant. There are two angular rates, and confusing them ruins the argument. Lowercase omega is the rapid spin about the top's own axle. Capital omega is the much slower precession of that axle around the vertical. The centre of mass is a distance ell from the support. Gravity acts downward there, so r cross m g supplies the sideways torque. The angle between the axle and the upward vertical is theta. First calculate the torque magnitude. It is the magnitude of r cross m g. The angle between r and downward gravity is pi minus theta, whose sine is the same as sine theta. Therefore the torque magnitude is m g ell sine theta. That sine factor has a direct geometric meaning. If the axle were vertical, gravity's line of action would pass through the support and the torque would vanish. Tilting the axle gives gravity a perpendicular lever arm ell sine theta. Now calculate how fast the tip of L moves in angular-momentum space. Its horizontal circle has radius L sine theta. A point moving around that circle at angular rate capital omega has speed capital omega times L sine theta. This is ordinary circular-motion geometry, but the circle lives in vector space. The radius is the horizontal component of L, and the speed around it is the magnitude of d L by d t. The dynamical law supplies that vector speed. Since d L by d t equals torque, the speed of the tip must equal the gravitational torque magnitude. Equate them. Capital omega times L sine theta equals m g ell sine theta. This line is the whole balance: the required turning rate of L on the left, and the turning rate supplied by gravity on the right. For a nonzero tilt, sine theta appears on both sides and cancels. The steady-precession rate is therefore m g ell divided by L. The units already make sense. Torque has units of angular momentum per time, so torque divided by L has units of one over time, exactly the units of an angular rate. For rapid spin about a symmetry axis, angular momentum is approximately moment of inertia I times spin rate lowercase omega. A heavier rim gives a larger I, and faster rotation gives a larger omega. Substitute that spin angular momentum. Capital omega is approximately m g ell divided by I times lowercase omega. This is the standard rate for steady precession of a fast symmetric top. The formula has a clean physical reading. More gravitational torque, from more weight or a longer lever arm, turns L faster. More angular momentum, from a larger moment of inertia or faster spin, makes the same torque turn the vector more slowly. Notice what did not survive the cancellation: the tilt angle. In this ideal steady, fast-top model, the leading precession rate is independent of theta. That does not say every real release is steady, or that a top may be tilted arbitrarily without complication. Friction slowly reduces the spin, large wobbling violates the steady approximation, and a badly asymmetric body needs a more complete treatment. Within the model we stated, however, the inverse dependence on spin is a sharp prediction that we can test."},{"title":"Faster Spin, Slower Precession","startSeconds":568.5615416666666,"narration":"The derived formula makes a prediction that initially sounds backward. Spin the top faster, and its slow motion around the vertical becomes slower, not faster. Let the two tops be identical, with the same mass, lever arm, and moment of inertia. Take m g ell divided by I to be twenty per second squared. Only the spin rate changes. The first top spins at one hundred radians per second, giving a precession rate of zero point two radians per second. The second spins twice as fast, at two hundred, and the formula predicts only zero point one. Run both predictions over the same interval. The original top's angular-momentum tip travels one full circuit. The faster-spinning top covers only half a circuit in that time. The pictures are not claiming that the faster top has less angular momentum. It has more. That larger vector is exactly why its direction changes by a smaller angle each second. Gravity has not become weaker, and the torque has not disappeared. For identical geometry it supplies the same sideways change d L in each second. The same change attached to a vector twice as large produces half the angular turn. This is the part that sounds backward only if we use the word rotation for two different things. The top spins rapidly about its own axle, while the axle precesses slowly around the vertical. Increasing the first rate decreases the second. The inverse relationship fits in one line. Capital omega is a fixed top-dependent constant divided by lowercase omega. Doubling lowercase omega doubles the spin angular momentum. Because the gravitational torque is essentially unchanged, it then takes twice as much time to turn L through a given angle. A practical comparison must be made before friction changes the spin very much, and with nearly equal tilt and geometry. Under those conditions, doubling the spin rate should approximately halve the steady-precession rate. As a real top loses spin to friction, the same formula predicts that its precession tends to speed up. Eventually the fast-top approximation fails, the motion becomes more complicated, and the top falls. The apparent stability was never freedom from gravity. It was gravity steering a large angular momentum slowly. A bicycle wheel suspended from one end makes that steering unusually easy to see."},{"title":"The Bicycle Wheel on a String","startSeconds":729.0074999999999,"narration":"The same argument becomes striking with a bicycle wheel. Tie a string to one end of the axle, hold the wheel with its axle horizontal, spin it, and release. The string supplies a fixed support point O while the wheel's centre of mass hangs well to one side. If the wheel is not spinning, there is no mystery. Gravity turns the axle downward about the supported end, and the wheel drops. Now remove that imagined fall and spin the wheel rapidly. Its angular momentum points along the axle. The direction follows the right-hand rule, so reversing the wheel's spin reverses L. Gravity still acts at C. The lever arm r runs from the string's support to the centre, so gravity produces r cross m g. With r horizontal and gravity downward, the torque points sideways. That sideways torque is perpendicular to the wheel's spin angular momentum. It therefore changes the direction of L rather than rapidly reducing its magnitude. The axle follows the turning vector. The unsupported centre does begin to accelerate, but not simply downward. As L turns sideways, the axle turns sideways with it. The continual response to gravity becomes motion around the vertical support line. The wheel therefore swings around the vertical line through O. This is the same precession as the top, now exposed in a form where the single support point and the long lever arm are easy to see. Write the chain once more. The gravitational torque is r cross m g. Torque is d L by d t. For a fast wheel in approximately steady precession, the angular rate is m g ell divided by L. The demonstration also checks direction. Reverse the spin and L reverses, while the gravitational torque for the same pose does not. The sideways change then carries the axle around the vertical in the opposite sense. Spin the wheel faster and L grows. The torque is almost unchanged, so the wheel precesses more slowly. Let friction reduce the spin and the precession generally quickens until the simple steady model ceases to describe the motion well. Real tops usually add a small bobbing of the tilt angle called nutation, caused by how they are released and by departures from perfectly steady precession. We can now answer the opening question without invoking a hidden supporting force. Gravity tries to turn the body about its contact or support point. Rapid spin supplies a large angular-momentum vector, and the gravitational torque changes that vector mainly sideways. The vector tip therefore travels around the vertical, the axle follows it, and we see precession. A top does not refuse gravity. Its spin changes the direction in which gravity first makes the motion develop. That is why a rapidly spinning top stays up for a while, why it circles slowly instead of simply falling, and why making it spin faster slows that circle down."}]}}