{"version":1,"lectureId":"01M14TYA5P7YWR2CMWJYKMN214","attempt":0,"publication":{"slug":"instantaneous-axis-of-rotation","title":"Kinematics of a Body: The Instantaneous Axis","subject":"engineering","summary":"A rigid body carries exactly one restriction: the distances between its points never change. We differentiate that restriction to obtain the velocity relation v_B = v_A + omega cross r, then watch the body move under its translation and rotation terms. A rolling wheel makes the instantaneous centre visible and shows why a different material point occupies it from one instant to the next. Finally, one continuous three-dimensional chapter turns the point at rest into a tilted instantaneous axis, sweeps the family of points along it, and follows a material point through Chasles' screw motion. Live limiting cases reduce the screw to pure rotation and pure translation.","metaDescription":"The rigid-body velocity relation, the instantaneous centre of a rolling wheel, and the screw axis that replaces it in three dimensions.","transcript":"Kinematics of a rigid body begins with one promise: distances inside the body never change. That promise will give us its complete velocity law. Here is the body itself, a collection of material points locked together. Pick any two of them and call them A and B. The vector r runs from A to B, and rigidity says its length never changes. Write that fixed length as r dotted with itself, equal to the constant L squared. Differentiate. The constant disappears, leaving r dotted with its own rate of change equal to zero. So that rate points square to r. Here is the whole family. Change the length and even reverse the arrow; it stays perpendicular. Every member can be written as omega crossed with r. For a rigid body, one angular velocity omega works for every pair of points at once. It belongs to the body, not to A or B. Now compose the positions. Start at the position of A, add r, and land at the position of B. Differentiate that addition. The first velocity is v A. The second term is the rate of r, which becomes omega cross r. Their sum is v B. A translates with v A while the body turns about A with angular velocity omega. B carries both effects, so its green arrow follows the sum we just built. Now let the body move through the translation and rotation we just composed. Translation plus rotation is the entire instantaneous freedom of a rigid body. Every construction that follows is this one relation read in a different way. In the plane, the rigid-body relation has a consequence you can watch. Here is a wheel rolling along the ground without slipping. The material point touching the ground is not sliding. At this instant C has velocity zero, so it is momentarily at rest. That is the instantaneous centre. Now take A in the relation to be C. Its velocity term is zero. What remains says that every point P moves as omega crossed with the vector from C to P. The hub is one radius from C, so it moves at omega R. The top material point is two radii away, so its speed is twice the hub speed. At Q, the same rule gives a velocity perpendicular to C Q. Its length grows in direct proportion to Q's distance from C. At one instant, then, the whole velocity field looks exactly like a wheel pinned at C and rotating about it. But C is not one fixed material point. Watch C zero leave the ground as the wheel rolls. A moment later, a different material point is touching down, and that new point is the one at rest. The same idea gives a ruler construction. Here is a moving bar with the velocity direction already attached at A and B. Each point circles C, so draw a perpendicular to each velocity. The centre must lie on both lines. The two lines meet here, at C. From that one point, every speed is omega times the distance out to the material point. There is one limiting case. If the two velocities are equal and parallel, their perpendiculars are parallel too. They never meet, so C is at infinity and the body is purely translating. Lift the body into three dimensions. This solid extends through all three coordinate directions, with its material points locked into one shape. Which of those points, if any, are momentarily at rest? Ask the rigid-body relation. Choose A on the body, draw its velocity, and keep the body's one angular velocity omega in view. We are hunting for a point P whose velocity is zero. The cross-product term must cancel v A exactly. It can do that only with the part square to omega, because every omega cross r is perpendicular to omega. Here the magenta arrow is that exact cancellation. For the moment v A is entirely square to omega. Solving the cancellation gives this particular displacement r zero. But r zero is only one answer. Add any multiple lambda omega and the cross product does not change. Watch lambda sweep through its values: every point it reaches is another solution, so the solutions fill a line. That tilted line is the instantaneous axis of rotation. Every green location on it has zero velocity at this instant. A point off the axis swings around it in a circle lying square to omega. Its speed is omega times its perpendicular distance from the axis. The same construction at B uses the same omega. Turn the view and the geometry separates cleanly: the red axis is one tilted line in space, and both green velocities stand square to it. Omega belongs to the whole body. A and B do not get different angular velocities. What changes from point to point is r, and therefore the cross-product contribution. Now remove the assumption we just used. If v A is not square to omega, split it into a perpendicular piece and a parallel piece. Let those two pieces land. Complete their parallelogram, and its green diagonal is the original velocity v A. The cross product can cancel the perpendicular piece, exactly as before. It can never touch the parallel piece, whose projection formula is this. So the axis still exists, but its points now slide along omega instead of resting. This yellow arrow is the velocity shared by every point on it. Take a point away from the axis. Its velocity is the sum of a turn around the axis and that same slide along it. Follow one material point. The combined motion traces this yellow helix, winding around the tilted axis while advancing along it. This is Chasles' theorem. At every instant, rigid-body motion is a screw: a rotation about an axis together with a translation along that axis. Watch the numbered point perform both parts. The slide per unit turn is the pitch. First limiting case: let the slide shrink to nothing. The advancing helix closes into a circle, the axis velocity vanishes, and the motion becomes pure rotation. Second limiting case: restore the slide and let omega shrink to nothing. The winding opens into a straight path, the red omega arrow disappears, and equal green velocities show pure translation. Put turn and slide back together. The tilted axis, the helix and the travelling point return as the one picture that contains the general case. Three ideas carry the lecture. Keep the moving screw beside them while we read the list. First, any two points share one omega, and their velocities differ by omega cross the vector between them. Second, plane motion has an instantaneous centre, but the material point occupying it changes as the body moves. Third, spatial motion has an instantaneous axis. Rotation about it plus translation along it is the general screw motion, every instant.","watch":{"version":1,"scenes":[{"title":"The Rigid Body Relation","start":0,"end":112.45983333333334,"objects":{"arrow_rate":"a Vector [magenta] labelled \"frac(dif arrow(r), dif t)\" drawn in plane (start=(((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1.…, end=((((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1…)","arrow_va":"an Arrow [green] labelled \"arrow(v)_A\" drawn in plane (start=(((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0.…, end=((((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0…)","arrow_vb":"an Arrow [green] labelled \"arrow(v)_B\" drawn in plane (start=(((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1.…, end=((((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1…)","body":"a Polygon [blue] drawn in plane (vertices=((((2.1 + (0.55 * motion)) + ((-1.2000000000000002 * cos((0.35 …, fill_opacity=0.12)","card":"a Title that says \"Engineering Dynamics — Kinematics of a Body: The Instantaneous Axis\"","curl_omega":"a CurvedArrow [red] labelled \"arrow(omega)\" drawn in plane (start=((((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0…, end=((((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0…, bend=0.45)","family":"a VariableNumber (initial_value=0.48, format_spec='.1f')","heading":"a Heading that says \"Two Points, One Angular Velocity\"","heading_sum":"a Heading that says \"Position Adds, So Velocity Adds\"","motion":"a VariableNumber (format_spec='.1f')","plane":"a Figure (x_range=(0.5, 6.0), y_range=(0.5, 4.8), aspect=(5.5, 4.3))","point":"a Point [yellow] drawn in plane (location=(2.1, 1.7))","point_2":"a Point [yellow] drawn in plane (location=(4.0, 3.3))","position_sum":"a Math [text] that says \"$arrow(x)_A + arrow(r) = arrow(x)_B$\"","pt_a":"a Point [text] labelled \"A\" drawn in plane (location=(((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0.…)","pt_b":"a Point [text] labelled \"B\" drawn in plane (location=(((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1.…)","rate_law":"a Math [text] that says \"$frac(dif arrow(r), dif t)$\"","relation":"a Math [text] that says \"$arrow(v)_B = arrow(v)_A + arrow(omega) times arrow(r)$\"","right_rate":"an Angle [yellow] drawn in plane (vertex=(((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1.…, sides=((((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0…, right_angle=True)","seg_r":"a Line [yellow] labelled \"arrow(r)\" drawn in plane (start=(((2.1 + (0.55 * motion)) + ((0.0 * cos((0.35 * motion))) - (0.…, end=(((2.1 + (0.55 * motion)) + ((1.9 * cos((0.35 * motion))) - (1.…)","velocity_sum":"a Math [text] that says \"$arrow(v)_A + arrow(omega) times arrow(r) = arrow(v)_B$\"","work":"a Derivation [text] that says \"$arrow(r) dot arrow(r) &= L^2 \\ 2 thin arrow(r) dot frac(dif arrow(r), dif t) &= 0$\""},"beats":[{"start":0,"say":"Kinematics of a rigid body begins with one promise: distances inside the body never change. 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The vector r runs from A to B, and rigidity says its length never changes.","live":null,"does":[[11.382,"heading is shown on the screen, written out."],[11.382,"plane is shown on the screen, written out."],[11.382,"body is shown on the screen, written out."],[13.366999999999999,"pt_a is shown on the screen, written out."],[13.366999999999999,"point is shown on the screen, grown."],[15.366999999999999,"point is hidden from the screen."],[18.604,"pt_b is shown on the screen, written out."],[18.604,"point_2 is shown on the screen, grown."],[19.869,"seg_r is shown on the screen, written out."],[20.604,"point_2 is hidden from the screen."]]},{"start":25.74,"say":"Write that fixed length as r dotted with itself, equal to the constant L squared.","live":["plane","heading","body","pt_a","pt_b","seg_r"],"does":[[26.087999999999997,"plane moves to a new place on the board."],[26.087999999999997,"work is shown on the screen, written out."],[28.038999999999998,"work (the \"arrow(r) dot arrow(r)\" part) is emphasized."],[30.93,"work (the \"L^2\" part) is emphasized."],[30.93,"work (the \"arrow(r) dot arrow(r)\" part) is no longer emphasized."],[31.9395,"work (the \"L^2\" part) is no longer emphasized."]]},{"start":32.5395,"say":"Differentiate. The constant disappears, leaving r dotted with its own rate of change equal to zero. So that rate points square to r.","live":null,"does":[[32.946,"work is shown on the screen, written out."],[36.697,"work (the \"arrow(r)\" part) is emphasized."],[37.916,"arrow_rate is shown on the screen, written out."],[37.916,"rate_law is shown on the screen, written out."],[37.916,"work (the \"arrow(r)\" part) is no longer emphasized."],[37.916,"work (the \"frac(dif arrow(r), dif t)\" part) is emphasized."],[41.456999999999994,"right_rate is shown on the screen, written out."],[42.8035,"work (the \"frac(dif arrow(r), dif t)\" part) is no longer emphasized."]]},{"start":43.403499999999994,"say":"Here is the whole family. Change the length and even reverse the arrow; it stays perpendicular. Every member can be written as omega crossed with r.","live":["rate_law","plane","heading","body","pt_a","pt_b","seg_r","arrow_rate","right_rate"],"does":[[43.403499999999994,"right_rate is hidden from the screen."],[45.574,"arrow_rate is redrawn as the numbers it depends on change."],[45.574,"family ticks to -0.58."],[51.34499999999999,"rate_law becomes \"$frac(dif arrow(r), dif t) = arrow(omega) times arrow(r)$\"."],[51.867,"rate_law (the \"arrow(omega)\" part) is emphasized."],[53.88699999999999,"rate_law (the \"arrow(omega)\" part) is no longer emphasized."]]},{"start":54.486999999999995,"say":"For a rigid body, one angular velocity omega works for every pair of points at once. It belongs to the body, not to A or B.","live":["rate_law","plane","heading","body","pt_a","pt_b","seg_r","arrow_rate"],"does":[[57.831,"rate_law (the \"arrow(omega)\" part) is emphasized."],[62.347,"rate_law (the \"arrow(omega)\" part) is no longer emphasized."],[64.40199999999999,"plane moves to a new place on the board."],[64.40199999999999,"arrow_rate is hidden from the screen."],[64.40199999999999,"heading is hidden from the screen — left the board."],[64.40199999999999,"rate_law is hidden from the screen — left the board."],[64.40199999999999,"work is hidden from the screen — left the board."]]},{"start":65.002,"say":"Now compose the positions. Start at the position of A, add r, and land at the position of B.","live":["plane","body","pt_a","pt_b","seg_r"],"does":[[65.002,"heading_sum is shown on the screen, written out."],[65.64,"position_sum is shown on the screen, written out."]]},{"start":72.759,"say":"Differentiate that addition. The first velocity is v A. The second term is the rate of r, which becomes omega cross r. Their sum is v B.","live":["plane","body","pt_a","pt_b","seg_r","position_sum","heading_sum"],"does":[[72.846,"velocity_sum is shown on the screen, written out."]]},{"start":83.791,"say":"A translates with v A while the body turns about A with angular velocity omega. B carries both effects, so its green arrow follows the sum we just built.","live":["plane","body","pt_a","pt_b","seg_r","position_sum","velocity_sum","heading_sum"],"does":[[84.093,"arrow_va is shown on the screen, written out."],[86.554,"curl_omega is shown on the screen, written out."],[90.769,"arrow_vb is shown on the screen, written out."]]},{"start":95.6295,"say":"Now let the body move through the translation and rotation we just composed.","live":["plane","body","pt_a","pt_b","seg_r","position_sum","velocity_sum","heading_sum","arrow_va","curl_omega","arrow_vb"],"does":[[96.82600000000001,"body is redrawn as the numbers it depends on change."],[96.82600000000001,"pt_a is redrawn as the numbers it depends on change."],[96.82600000000001,"pt_b is redrawn as the numbers it depends on change."],[96.82600000000001,"seg_r is redrawn as the numbers it depends on change."],[96.82600000000001,"arrow_va is redrawn as the numbers it depends on change."],[96.82600000000001,"curl_omega is redrawn as the numbers it depends on change."],[96.82600000000001,"arrow_vb is redrawn as the numbers it depends on change."],[96.82600000000001,"motion ticks to 1.2."]]},{"start":100.8505,"say":"Translation plus rotation is the entire instantaneous freedom of a rigid body. 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Erect a perpendicular to each velocity. Their intersection is $C$; parallel lines place $C$ at infinity. Every point has speed $omega d$, where $d$ is its distance from $C$.\"","spoke_cq":"a Line [gray] drawn in wheel (start=((3.0 + (1.4 * roll)), 0.6), end=(((3.0 + (1.4 * roll)) + (1.4 * cos((0.7853981633974483 - roll)…, dashed=True)","spoke_ct":"a Line [gray] drawn in wheel (start=((3.0 + (1.4 * roll)), 0.6), end=(((3.0 + (1.4 * roll)) + (1.4 * cos((1.5707963267948966 - roll)…, dashed=True)","wheel":"a Figure (x_range=(0.5, 6.9), y_range=(0.2, 4.1), aspect=(6.4, 3.9))"},"beats":[{"start":112.45983333333334,"say":"In the plane, the rigid-body relation has a consequence you can watch. Here is a wheel rolling along the ground without slipping.","live":[],"does":[[112.45983333333334,"wheel is shown on the screen, written out."],[112.45983333333334,"ground is shown on the screen, written out."],[112.45983333333334,"rim is shown on the screen, written out."],[112.45983333333334,"pt_hub is shown on the screen, written out."],[112.76183333333334,"heading_ic is shown on the screen, written out."]]},{"start":121.09433333333334,"say":"The material point touching the ground is not sliding. At this instant C has velocity zero, so it is momentarily at rest. That is the instantaneous centre.","live":["wheel","heading_ic","ground","rim","pt_hub"],"does":[[122.21983333333334,"pt_c is shown on the screen, written out."],[125.74983333333333,"point is shown on the screen, grown."],[127.74983333333333,"point is hidden from the screen."],[128.14183333333332,"concept_ic is shown on the screen, written out."]]},{"start":132.58433333333335,"say":"Now take A in the relation to be C. 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Its length grows in direct proportion to Q's distance from C.","live":["concept_ic","formula_ic","wheel","heading_ic","ground","rim","pt_hub","pt_c","spoke_ct","arrow_hub","pt_top","arrow_top"],"does":[[155.35683333333333,"pt_q is shown on the screen, written out."],[155.35683333333333,"point_2 is shown on the screen, grown."],[156.70283333333333,"arrow_q is shown on the screen, written out."],[157.29483333333334,"right_q is shown on the screen, written out."],[157.35683333333333,"point_2 is hidden from the screen."],[158.39783333333332,"spoke_cq is shown on the screen, written out."],[160.03483333333332,"arrow_q is indicated — a transient flash."]]},{"start":164.24533333333335,"say":"At one instant, then, the whole velocity field looks exactly like a wheel pinned at C and rotating about it.","live":["concept_ic","formula_ic","wheel","heading_ic","ground","rim","pt_hub","pt_c","spoke_ct","arrow_hub","pt_top","arrow_top","pt_q","spoke_cq","arrow_q","right_q"],"does":[[168.45983333333334,"concept_ic (the \"pinned there\" part) is emphasized."],[170.66583333333332,"concept_ic (the \"pinned there\" part) is no longer emphasized."]]},{"start":171.26583333333332,"say":"But C is not one fixed material point. Watch C zero leave the ground as the wheel rolls. A moment later, a different material point is touching down, and that new point is the one at rest.","live":null,"does":[[177.63983333333334,"rim is redrawn as the numbers it depends on change."],[177.63983333333334,"pt_hub is redrawn as the numbers it depends on change."],[177.63983333333334,"pt_c is redrawn as the numbers it depends on change."],[177.63983333333334,"spoke_ct is redrawn as the numbers it depends on change."],[177.63983333333334,"arrow_hub is redrawn as the numbers it depends on change."],[177.63983333333334,"pt_top is redrawn as the numbers it depends on change."],[177.63983333333334,"arrow_top is redrawn as the numbers it depends on change."],[177.63983333333334,"pt_q is redrawn as the numbers it depends on change."],[177.63983333333334,"spoke_cq is redrawn as the numbers it depends on change."],[177.63983333333334,"arrow_q is redrawn as the numbers it depends on change."],[177.63983333333334,"right_q is redrawn as the numbers it depends on change."],[177.63983333333334,"pt_next is redrawn as the numbers it depends on change."],[177.63983333333334,"roll ticks to 0.75."],[180.14683333333335,"pt_next is shown on the screen, written out."],[180.14683333333335,"point_3 is shown on the screen, grown."],[182.14683333333335,"point_3 is hidden from the screen."],[184.79133333333334,"concept_ic is hidden from the screen — left the board."],[184.79133333333334,"formula_ic is hidden from the screen — left the board."],[184.79133333333334,"heading_ic is hidden from the screen — left the board."],[184.79133333333334,"wheel is hidden from the screen — left the board."],[184.79133333333334,"ground is hidden from the screen — wheel left the board."],[184.79133333333334,"rim is hidden from the screen — wheel left the board."],[184.79133333333334,"pt_hub is hidden from the screen — wheel left the board."],[184.79133333333334,"pt_c is hidden from the screen — wheel left the board."],[184.79133333333334,"spoke_ct is hidden from the screen — wheel left the board."],[184.79133333333334,"arrow_hub is hidden from the screen — wheel left the board."],[184.79133333333334,"pt_top is hidden from the screen — wheel left the board."],[184.79133333333334,"arrow_top is hidden from the screen — wheel left the board."],[184.79133333333334,"pt_q is hidden from the screen — wheel left the board."],[184.79133333333334,"spoke_cq is hidden from the screen — wheel left the board."],[184.79133333333334,"arrow_q is hidden from the screen — wheel left the board."],[184.79133333333334,"right_q is hidden from the screen — wheel left the board."],[184.79133333333334,"pt_next is hidden from the screen — wheel left the board."]]},{"start":185.99133333333333,"say":"The same idea gives a ruler construction. Here is a moving bar with the velocity direction already attached at A and B.","live":[],"does":[[185.99133333333333,"heading_build is shown on the screen, written out."],[185.99133333333333,"build is shown on the screen, written out."],[185.99133333333333,"rod is shown on the screen, written out."],[185.99133333333333,"pt_ba is shown on the screen, written out."],[185.99133333333333,"pt_bb is shown on the screen, written out."],[185.99133333333333,"arrow_bva is shown on the screen, written out."],[185.99133333333333,"arrow_bvb is shown on the screen, written out."],[187.81383333333332,"build moves to a new place on the board."],[187.81383333333332,"rules is shown on the screen, written out."],[191.27383333333333,"rules (the \"velocity direction\" part) is emphasized."]]},{"start":194.61383333333333,"say":"Each point circles C, so draw a perpendicular to each velocity. The centre must lie on both lines.","live":["rules","build","heading_build","rod","pt_ba","pt_bb","arrow_bva","arrow_bvb"],"does":[[194.96183333333335,"perp_b is shown on the screen, drawn."],[194.96183333333335,"right_a is shown on the screen, written out."],[194.96183333333335,"right_b is shown on the screen, written out."],[197.27283333333332,"perp_a is shown on the screen, drawn."],[197.27283333333332,"rules (the \"perpendicular\" part) is emphasized."],[197.27283333333332,"rules (the \"velocity direction\" part) is no longer emphasized."]]},{"start":202.86533333333333,"say":"The two lines meet here, at C. From that one point, every speed is omega times the distance out to the material point.","live":["rules","build","heading_build","rod","pt_ba","pt_bb","arrow_bva","arrow_bvb","perp_a","perp_b","right_a","right_b"],"does":[[203.90983333333332,"rules (the \"intersection\" part) is emphasized."],[203.90983333333332,"rules (the \"perpendicular\" part) is no longer emphasized."],[204.11883333333333,"pt_bc is shown on the screen, written out."],[204.8268333333333,"point_4 is shown on the screen, grown."],[206.8268333333333,"point_4 is hidden from the screen."],[208.1128333333333,"rules (the \"$omega d$\" part) is emphasized."],[208.1128333333333,"rules (the \"intersection\" part) is no longer emphasized."]]},{"start":211.7663333333333,"say":"There is one limiting case. If the two velocities are equal and parallel, their perpendiculars are parallel too. They never meet, so C is at infinity and the body is purely translating.","live":["rules","build","heading_build","rod","pt_ba","pt_bb","arrow_bva","arrow_bvb","perp_a","perp_b","right_a","right_b","pt_bc"],"does":[[216.0388333333333,"arrow_bva is hidden from the screen."],[216.0388333333333,"arrow_bvb is hidden from the screen."],[216.0388333333333,"perp_a is hidden from the screen."],[216.0388333333333,"perp_b is hidden from the screen."],[216.0388333333333,"right_a is hidden from the screen."],[216.0388333333333,"right_b is hidden from the screen."],[216.0388333333333,"pt_bc is hidden from the screen."],[216.0388333333333,"par_a is shown on the screen, written out."],[216.51483333333329,"par_b is shown on the screen, written out."],[217.7578333333333,"par_perp_a is shown on the screen, drawn."],[218.60483333333332,"par_perp_b is shown on the screen, drawn."],[220.4278333333333,"rules (the \"$omega d$\" part) is no longer emphasized."],[220.4278333333333,"rules (the \"parallel lines\" part) is emphasized."],[224.934125,"rules (the \"parallel lines\" part) is no longer emphasized."],[225.184125,"build is hidden from the screen — left the board."],[225.184125,"rod is hidden from the screen — build left the board."],[225.184125,"pt_ba is hidden from the screen — build left the board."],[225.184125,"pt_bb is hidden from the screen — build left the board."],[225.184125,"par_a is hidden from the screen — build left the board."],[225.184125,"par_b is hidden from the screen — build left the board."],[225.184125,"par_perp_a is hidden from the screen — build left the board."],[225.184125,"par_perp_b is hidden from the screen — build left the board."],[225.184125,"heading_build is hidden from the screen — left the board."],[225.184125,"rules is hidden from the screen — left the board."]]}]},{"title":"The Instantaneous Axis and Screw Motion","start":226.22579166666668,"end":481.38031249999995,"objects":{"answer_axis":"a Math [text] that says \"$arrow(r) = arrow(r)_0 + lambda thin arrow(omega)$\"","arrow_a":"a Vector [green] labelled \"arrow(v)_A\" drawn in frame (start=(1.2, -0.8, 0.2), end=(1.48026, -0.30752, -0.1159))","arrow_b":"a Vector [green] labelled \"arrow(v)_B\" drawn in frame (start=(1.68, -0.56, 1.0), end=(1.96026, -0.06752000000000002, 0.6840999999999999))","arrow_parallel":"a Vector [yellow] labelled \"arrow(v)_(parallel)\" drawn in frame (start=(1.2, -0.8, 0.2), end=(1.6185492184628494, -0.5907253907685753, 0.8975820307714157))","arrow_perp":"a Vector [blue] labelled \"arrow(v)_(bot)\" drawn in frame (start=(1.2, -0.8, 0.2), end=(1.48026, -0.30752, -0.1159))","arrow_total":"a Vector [green] labelled \"arrow(v)_A\" drawn in frame (start=(1.2, -0.8, 0.2), end=(1.8988092184628493, -0.09824539076857527, 0.5816820307714158))","axis":"a Line [red] labelled \"upright(\"instantaneous axis\")\" drawn in frame (start=(-0.45000000000000007, -0.45500000000000007, -0.726), end=(1.71, 0.6249999999999999, 2.874), dashed=True)","axis_lambda":"a VariableNumber (initial_value=-0.75, format_spec='.1f')","axis_velocity":"a Vector [yellow] labelled \"arrow(v)_(parallel)\" drawn in frame (start=((0.29999999999999993 + (axis_lambda * 0.6)), (-0.0800000000000…, end=(((0.29999999999999993 + (axis_lambda * 0.6)) + 0.4185492184628…)","body":"a Cylinder [blue] drawn in frame (start=(1.77, -1.72, 1.99), end=(1.11, 0.36, -0.79), radius=0.6)","circle_a":"a Circle [gray] drawn in frame (center=(0.29999999999999993, -0.08000000000000007, 0.524), radius=1.197236818678744, normal_vector=(0.6, 0.3, 1.0))","cross_a":"a Vector [magenta] labelled \"arrow(omega) times arrow(r)_0\" drawn in frame (start=(1.2, -0.8, 0.2), end=(0.91974, -1.29248, 0.5159))","frame":"an Axes3D (x_range=(-2.4, 2.4), y_range=(-2.4, 2.4), z_range=(-1.2, 3.6))","heading_recap":"a Heading that says \"What to Carry Away\"","heading_screw":"a Heading that says \"When Nothing Is at Rest\"","helix":"a ParametricCurve [yellow] labelled \"upright(\"screw path\")\" drawn in frame (function=<function>, t_range=(0.0, 6.283185307179586))","helix_rider":"a Point [green] labelled \"0.0\" drawn in frame (location=(((0.29999999999999993 + ((0.2 + (0.28 * helix_t)) * 0.49827287…)","helix_t":"a VariableNumber (format_spec='.1f')","omega_vec":"a Vector [red] labelled \"arrow(omega)\" drawn in frame (start=(-1.9, 0.98, 2.01), end=(-1.1775043252724622, 1.3412478373637688, 3.214159457879229))","pitch":"a Math [text] that says \"$h = frac(v_(parallel), omega)$\"","point":"a Point [yellow] drawn in frame (location=(1.2, -0.8, 0.2))","pt_a":"a Point [text] labelled \"A\" drawn in frame (location=(1.2, -0.8, 0.2))","pt_b":"a Point [text] labelled \"B\" drawn in frame (location=(1.68, -0.56, 1.0))","question":"a Text [text] that says \"In three dimensions, which points of the body are momentarily at rest?\"","r0_vector":"a Vector [yellow] labelled \"arrow(r)_0\" drawn in frame (start=(0.29999999999999993, -0.08000000000000007, 0.524), end=(1.2, -0.8, 0.2))","recap":"a Block [text] that says \"Two points share one $arrow(omega)$: $arrow(v)_B = arrow(v)_A + arrow(omega) times arrow(r)$. Plane motion has a moving instantaneous centre. Spatial motion has an instantaneous axis; turn plus slide gives a screw.\"","rest_law":"a Math [text] that says \"$arrow(v)_(C(lambda)) = 0$\"","rest_point":"a Point [green] labelled \"-0.8\" drawn in frame (location=((0.29999999999999993 + (axis_lambda * 0.6)), (-0.0800000000000…)","right_cross":"an Angle [yellow] drawn in frame (vertex=(1.2, -0.8, 0.2), sides=((1.5487910153857078, -0.6256044923071461, 0.7813183589761798),…, right_angle=True)","slide_only":"a Line [yellow] drawn in frame (start=(0.3996545758244879, -0.03017271208775609, 0.69009095970748), end=(1.2762620090835795, 0.4081310045417897, 2.1511033484726325))","spoke_a":"a Line [gray] drawn in frame (start=(0.29999999999999993, -0.08000000000000007, 0.524), end=(1.2, -0.8, 0.2), dashed=True)","sum_side_parallel":"a Line [gray] drawn in frame (start=(1.6185492184628494, -0.5907253907685753, 0.8975820307714157), end=(1.8988092184628493, -0.09824539076857527, 0.5816820307714158), dashed=True)","sum_side_perp":"a Line [gray] drawn in frame (start=(1.48026, -0.30752, -0.1159), end=(1.8988092184628493, -0.09824539076857527, 0.5816820307714158), dashed=True)","translation_a":"a Vector [green] labelled \"arrow(v)\" drawn in frame (start=(1.2, -0.8, 0.2), end=(1.6185492184628494, -0.5907253907685753, 0.8975820307714157))","translation_b":"a Vector [green] labelled \"arrow(v)\" drawn in frame (start=(1.68, -0.56, 1.0), end=(2.0985492184628494, -0.35072539076857534, 1.6975820307714158))","turn_only":"a Circle [yellow] drawn in frame (center=(0.8379582924540336, 0.18897914622701678, 1.4205971540900562), radius=0.72, normal_vector=(0.6, 0.3, 1.0))","work_axis":"a Derivation [text] that says \"$arrow(v)_P &= arrow(v)_A + arrow(omega) times arrow(r) \\ 0 &= arrow(v)_A + arrow(omega) times arrow(r) \\ arrow(r)_0 &= frac(arrow(omega) times arrow(v)_A, arrow(omega) dot arrow(omega))$\"","work_screw":"a Derivation [text] that says \"$arrow(v)_A &= arrow(v)_(bot) + arrow(v)_(parallel) \\ arrow(v)_(parallel) &= frac(arrow(v)_A dot arrow(omega), arrow(omega) dot arrow(omega)) thin arrow(omega) \\ arrow(v)_P &= arrow(v)_(parallel) + arrow(omega) times arrow(r)$\""},"beats":[{"start":226.22579166666668,"say":"Lift the body into three dimensions. This solid extends through all three coordinate directions, with its material points locked into one shape. Which of those points, if any, are momentarily at rest?","live":[],"does":[[226.22579166666668,"question is shown on the screen, written out."],[226.22579166666668,"frame is shown on the screen, written out."],[226.22579166666668,"body is shown on the screen, written out."],[229.4537916666667,"frame turns in its own slot."]]},{"start":240.2312916666667,"say":"Ask the rigid-body relation. Choose A on the body, draw its velocity, and keep the body's one angular velocity omega in view. We are hunting for a point P whose velocity is zero.","live":["frame","question","body"],"does":[[241.71679166666667,"frame moves to a new place on the board."],[241.71679166666667,"work_axis is shown on the screen, written out."],[243.4357916666667,"pt_a is shown on the screen, written out."],[243.4357916666667,"point is shown on the screen, grown."],[245.10779166666669,"arrow_a is shown on the screen, written out."],[245.4357916666667,"point is hidden from the screen."],[248.3117916666667,"omega_vec is shown on the screen, written out."],[252.60779166666669,"work_axis is shown on the screen, written out."]]},{"start":254.1592916666667,"say":"The cross-product term must cancel v A exactly. It can do that only with the part square to omega, because every omega cross r is perpendicular to omega. Here the magenta arrow is that exact cancellation.","live":["frame","question","body","pt_a","arrow_a","omega_vec"],"does":[[254.72879166666667,"work_axis (the \"arrow(omega) times arrow(r)\" part) is emphasized."],[256.39979166666666,"work_axis (the \"arrow(omega) times arrow(r)\" part) is no longer emphasized."],[256.39979166666666,"work_axis (the \"arrow(v)_A\" part) is emphasized."],[263.0527916666667,"right_cross is shown on the screen, written out."],[265.5957916666667,"cross_a is shown on the screen, written out."],[267.3717916666667,"arrow_a is indicated — a transient flash."],[268.5447916666667,"work_axis (the \"arrow(v)_A\" part) is no longer emphasized."]]},{"start":269.14479166666666,"say":"For the moment v A is entirely square to omega. Solving the cancellation gives this particular displacement r zero.","live":["frame","question","body","pt_a","arrow_a","omega_vec","cross_a","right_cross"],"does":[[269.14479166666666,"right_cross is hidden from the screen."],[273.2777916666667,"work_axis is shown on the screen, written out."],[275.0427916666667,"work_axis (the \"arrow(omega) times arrow(v)_A\" part) is emphasized."],[276.29679166666665,"r0_vector is shown on the screen, written out."],[277.30629166666665,"work_axis (the \"arrow(omega) times arrow(v)_A\" part) is no longer emphasized."]]},{"start":277.9062916666667,"say":"But r zero is only one answer. Add any multiple lambda omega and the cross product does not change. Watch lambda sweep through its values: every point it reaches is another solution, so the solutions fill a line.","live":["frame","question","body","pt_a","arrow_a","omega_vec","cross_a","r0_vector"],"does":[[277.9062916666667,"answer_axis is shown on the screen, written out."],[279.4967916666667,"answer_axis (the \"arrow(r)_0\" part) is emphasized."],[281.9347916666667,"rest_point is shown on the screen, written out."],[281.9347916666667,"answer_axis (the \"arrow(r)_0\" part) is no longer emphasized."],[281.9347916666667,"answer_axis (the \"lambda thin arrow(omega)\" part) is emphasized."],[286.6607916666667,"rest_point is redrawn as the numbers it depends on change."],[286.6607916666667,"axis_velocity is redrawn as the numbers it depends on change."],[286.6607916666667,"axis_lambda ticks to 1.55."],[292.08279166666665,"axis is shown on the screen, drawn."],[292.8022916666667,"answer_axis (the \"lambda thin arrow(omega)\" part) is no longer emphasized."]]},{"start":293.40229166666666,"say":"That tilted line is the instantaneous axis of rotation. Every green location on it has zero velocity at this instant.","live":["frame","question","body","answer_axis","pt_a","arrow_a","omega_vec","cross_a","r0_vector","axis","rest_point"],"does":[[294.42379166666666,"axis is indicated — a transient flash."],[299.36979166666663,"rest_law is shown on the screen, written out."],[299.36979166666663,"rest_law (the \"0\" part) is emphasized."],[301.39029166666666,"rest_law (the \"0\" part) is no longer emphasized."]]},{"start":301.9902916666667,"say":"A point off the axis swings around it in a circle lying square to omega. Its speed is omega times its perpendicular distance from the axis. The same construction at B uses the same omega.","live":["frame","question","body","answer_axis","rest_law","pt_a","arrow_a","omega_vec","cross_a","r0_vector","axis","rest_point"],"does":[[304.67179166666665,"circle_a is shown on the screen, drawn."],[307.5047916666667,"arrow_a is indicated — a transient flash."],[310.2687916666667,"spoke_a is shown on the screen, written out."],[312.39279166666665,"arrow_b is shown on the screen, written out."],[313.40279166666664,"pt_b is shown on the screen, written out."]]},{"start":316.0112916666667,"say":"Turn the view and the geometry separates cleanly: the red axis is one tilted line in space, and both green velocities stand square to it.","live":["frame","question","body","answer_axis","rest_law","pt_a","arrow_a","omega_vec","cross_a","r0_vector","axis","rest_point","spoke_a","circle_a","pt_b","arrow_b"],"does":[[316.2667916666667,"frame turns in its own slot."]]},{"start":326.3287916666667,"say":"Omega belongs to the whole body. A and B do not get different angular velocities. What changes from point to point is r, and therefore the cross-product contribution.","live":null,"does":[[329.48679166666665,"pt_a is indicated — a transient flash."],[329.90479166666665,"pt_b is indicated — a transient flash."],[333.81679166666663,"work_axis (the \"arrow(v)_A\" part) is emphasized."],[336.2087916666667,"work_axis (the \"arrow(omega) times arrow(r)\" part) is emphasized."],[336.2087916666667,"work_axis (the \"arrow(v)_A\" part) is no longer emphasized."],[338.1942916666667,"frame moves to a new place on the board."],[338.1942916666667,"rest_point is hidden from the screen."],[338.1942916666667,"answer_axis is hidden from the screen — left the board."],[338.1942916666667,"question is hidden from the screen — left the board."],[338.1942916666667,"rest_law is hidden from the screen — left the board."],[338.1942916666667,"work_axis is hidden from the screen — left the board."],[338.1942916666667,"work_axis (the \"arrow(omega) times arrow(r)\" part) is no longer emphasized."]]},{"start":339.3942916666667,"say":"Now remove the assumption we just used. If v A is not square to omega, split it into a perpendicular piece and a parallel piece.","live":["frame","body","pt_a","arrow_a","omega_vec","cross_a","r0_vector","axis","spoke_a","circle_a","pt_b","arrow_b"],"does":[[339.3942916666667,"heading_screw is shown on the screen, written out."],[339.3942916666667,"arrow_a is hidden from the screen."],[339.3942916666667,"cross_a is hidden from the screen."],[339.3942916666667,"r0_vector is hidden from the screen."],[339.3942916666667,"circle_a is hidden from the screen."],[339.3942916666667,"spoke_a is hidden from the screen."],[339.3942916666667,"arrow_b is hidden from the screen."],[339.3942916666667,"pt_b is hidden from the screen."],[345.00179166666663,"work_screw is shown on the screen, written out."],[345.83779166666665,"arrow_perp is shown on the screen, written out."],[347.11479166666663,"arrow_parallel is shown on the screen, written out."]]},{"start":348.9802916666667,"say":"Let those two pieces land. Complete their parallelogram, and its green diagonal is the original velocity v A.","live":["frame","body","pt_a","omega_vec","axis","heading_screw","arrow_perp","arrow_parallel"],"does":[[351.17479166666664,"sum_side_perp is shown on the screen, drawn."],[351.82479166666667,"sum_side_parallel is shown on the screen, drawn."],[353.57779166666666,"arrow_total is shown on the screen, written out."]]},{"start":357.08029166666665,"say":"The cross product can cancel the perpendicular piece, exactly as before. It can never touch the parallel piece, whose projection formula is this.","live":["frame","body","pt_a","omega_vec","axis","heading_screw","arrow_perp","arrow_parallel","sum_side_perp","sum_side_parallel","arrow_total"],"does":[[359.07779166666666,"work_screw (the \"arrow(v)_(bot)\" part) is emphasized."],[363.02479166666666,"work_screw (the \"arrow(v)_(bot)\" part) is no longer emphasized."],[363.02479166666666,"work_screw (the \"arrow(v)_(parallel)\" part) is emphasized."],[364.26779166666665,"work_screw is shown on the screen, written out."],[366.09029166666664,"work_screw (the \"arrow(v)_(parallel)\" part) is no longer emphasized."]]},{"start":366.69029166666667,"say":"So the axis still exists, but its points now slide along omega instead of resting. This yellow arrow is the velocity shared by every point on it.","live":null,"does":[[369.9637916666666,"axis_velocity is shown on the screen, written out."],[373.40079166666663,"work_screw is shown on the screen, written out."]]},{"start":377.3912916666667,"say":"Take a point away from the axis. Its velocity is the sum of a turn around the axis and that same slide along it.","live":["frame","body","pt_a","omega_vec","axis","heading_screw","arrow_perp","arrow_parallel","sum_side_perp","sum_side_parallel","arrow_total","axis_velocity"],"does":[[381.14079166666664,"work_screw (the \"arrow(omega) times arrow(r)\" part) is emphasized."],[382.90579166666663,"work_screw (the \"arrow(omega) times arrow(r)\" part) is no longer emphasized."],[382.90579166666663,"work_screw (the \"arrow(v)_(parallel)\" part) is emphasized."],[384.3572916666667,"work_screw (the \"arrow(v)_(parallel)\" part) is no longer emphasized."]]},{"start":384.95729166666666,"say":"Follow one material point. The combined motion traces this yellow helix, winding around the tilted axis while advancing along it.","live":null,"does":[[386.25779166666666,"helix_rider is shown on the screen, written out."],[389.64779166666665,"helix is shown on the screen, drawn."]]},{"start":394.20729166666666,"say":"This is Chasles' theorem. At every instant, rigid-body motion is a screw: a rotation about an axis together with a translation along that axis. Watch the numbered point perform both parts. The slide per unit turn is the pitch.","live":["frame","body","pt_a","omega_vec","axis","heading_screw","arrow_perp","arrow_parallel","sum_side_perp","sum_side_parallel","arrow_total","axis_velocity","helix","helix_rider"],"does":[[399.62879166666664,"helix_rider is redrawn as the numbers it depends on change."],[399.62879166666664,"helix_t ticks to 6.1."],[408.5917916666666,"pitch is shown on the screen, written out."],[409.40479166666665,"A box is drawn around pitch."]]},{"start":410.0047916666667,"say":"First limiting case: let the slide shrink to nothing. The advancing helix closes into a circle, the axis velocity vanishes, and the motion becomes pure rotation.","live":["frame","body","pt_a","omega_vec","axis","pitch","heading_screw","arrow_perp","arrow_parallel","sum_side_perp","sum_side_parallel","arrow_total","axis_velocity","helix","helix_rider"],"does":[[412.3847916666666,"arrow_perp is hidden from the screen."],[412.3847916666666,"arrow_parallel is hidden from the screen."],[412.3847916666666,"sum_side_perp is hidden from the screen."],[412.3847916666666,"sum_side_parallel is hidden from the screen."],[412.3847916666666,"arrow_total is hidden from the screen."],[414.7647916666666,"helix is hidden from the screen."],[414.7647916666666,"helix_rider is hidden from the screen."],[415.48479166666664,"turn_only is shown on the screen, drawn."],[417.40079166666663,"axis_velocity is hidden from the screen."]]},{"start":421.3092916666667,"say":"Second limiting case: restore the slide and let omega shrink to nothing. The winding opens into a straight path, the red omega arrow disappears, and equal green velocities show pure translation.","live":["frame","body","pt_a","omega_vec","axis","pitch","heading_screw","turn_only"],"does":[[427.6947916666666,"turn_only is hidden from the screen."],[428.47279166666664,"slide_only is shown on the screen, drawn."],[430.78279166666664,"omega_vec is hidden from the screen."],[430.78279166666664,"axis is hidden from the screen."],[432.43179166666664,"translation_a is shown on the screen, written out."],[432.70979166666666,"translation_b is shown on the screen, written out."]]},{"start":436.32479166666667,"say":"Put turn and slide back together. The tilted axis, the helix and the travelling point return as the one picture that contains the general case.","live":["frame","body","pt_a","pitch","heading_screw","slide_only","translation_a","translation_b"],"does":[[436.88279166666666,"omega_vec is shown on the screen, written out."],[437.8457916666666,"helix_t ticks to 0.2."],[438.10179166666666,"slide_only is hidden from the screen."],[438.10179166666666,"translation_a is hidden from the screen."],[438.10179166666666,"translation_b is hidden from the screen."],[439.9477916666666,"axis is shown on the screen, drawn."],[440.88779166666666,"helix is shown on the screen, drawn."],[442.29279166666663,"helix_rider is shown on the screen, written out."],[446.22879166666667,"heading_screw is hidden from the screen — left the board."],[446.22879166666667,"pitch is hidden from the screen — left the board."],[446.22879166666667,"work_screw is hidden from the screen — left the board."]]},{"start":447.42879166666665,"say":"Three ideas carry the lecture. Keep the moving screw beside them while we read the list.","live":["frame","body","pt_a","omega_vec","axis","helix","helix_rider"],"does":[[447.42879166666665,"heading_recap is shown on the screen, written out."],[447.77679166666667,"recap is shown on the screen, written out."],[450.2497916666666,"helix_rider is redrawn as the numbers it depends on change."],[450.2497916666666,"helix_t ticks to 5.9."]]},{"start":453.6827916666666,"say":"First, any two points share one omega, and their velocities differ by omega cross the vector between them.","live":["frame","body","pt_a","omega_vec","axis","helix","helix_rider","recap","heading_recap"],"does":[[454.03079166666663,"recap (the \"Two points\" part) is emphasized."],[455.99379166666665,"recap (the \"Two points\" part) is no longer emphasized."],[455.99379166666665,"recap (the \"one $arrow(omega)$\" part) is emphasized."]]},{"start":461.64379166666663,"say":"Second, plane motion has an instantaneous centre, but the material point occupying it changes as the body moves.","live":null,"does":[[462.11979166666663,"recap (the \"Plane motion\" part) is emphasized."],[462.11979166666663,"recap (the \"one $arrow(omega)$\" part) is no longer emphasized."]]},{"start":470.05729166666663,"say":"Third, spatial motion has an instantaneous axis. Rotation about it plus translation along it is the general screw motion, every instant.","live":null,"does":[[470.40579166666663,"recap (the \"Plane motion\" part) is no longer emphasized."],[470.40579166666663,"recap (the \"Spatial motion\" part) is emphasized."],[474.6777916666666,"recap (the \"Spatial motion\" part) is no longer emphasized."],[474.6777916666666,"recap (the \"turn plus slide\" part) is emphasized."],[480.0886458333333,"recap (the \"turn plus slide\" part) is no longer emphasized."],[480.3386458333333,"frame is hidden from the screen — left the board."],[480.3386458333333,"body is hidden from the screen — frame left the board."],[480.3386458333333,"pt_a is hidden from the screen — frame left the board."],[480.3386458333333,"omega_vec is hidden from the screen — frame left the board."],[480.3386458333333,"axis is hidden from the screen — frame left the board."],[480.3386458333333,"helix is hidden from the screen — frame left the board."],[480.3386458333333,"helix_rider is hidden from the screen — frame left the board."],[480.3386458333333,"heading_recap is hidden from the screen — left the board."],[480.3386458333333,"recap is hidden from the screen — left the board."]]}]}]},"durationSeconds":481,"chapters":[{"title":"The Rigid Body Relation","startSeconds":0,"narration":"Kinematics of a rigid body begins with one promise: distances inside the body never change. That promise will give us its complete velocity law. Here is the body itself, a collection of material points locked together. Pick any two of them and call them A and B. The vector r runs from A to B, and rigidity says its length never changes. Write that fixed length as r dotted with itself, equal to the constant L squared. Differentiate. The constant disappears, leaving r dotted with its own rate of change equal to zero. So that rate points square to r. Here is the whole family. Change the length and even reverse the arrow; it stays perpendicular. Every member can be written as omega crossed with r. For a rigid body, one angular velocity omega works for every pair of points at once. It belongs to the body, not to A or B. Now compose the positions. Start at the position of A, add r, and land at the position of B. Differentiate that addition. The first velocity is v A. The second term is the rate of r, which becomes omega cross r. Their sum is v B. A translates with v A while the body turns about A with angular velocity omega. B carries both effects, so its green arrow follows the sum we just built. Now let the body move through the translation and rotation we just composed. Translation plus rotation is the entire instantaneous freedom of a rigid body. Every construction that follows is this one relation read in a different way."},{"title":"The Instantaneous Centre","startSeconds":112.45983333333334,"narration":"In the plane, the rigid-body relation has a consequence you can watch. Here is a wheel rolling along the ground without slipping. The material point touching the ground is not sliding. At this instant C has velocity zero, so it is momentarily at rest. That is the instantaneous centre. Now take A in the relation to be C. Its velocity term is zero. What remains says that every point P moves as omega crossed with the vector from C to P. The hub is one radius from C, so it moves at omega R. The top material point is two radii away, so its speed is twice the hub speed. At Q, the same rule gives a velocity perpendicular to C Q. Its length grows in direct proportion to Q's distance from C. At one instant, then, the whole velocity field looks exactly like a wheel pinned at C and rotating about it. But C is not one fixed material point. Watch C zero leave the ground as the wheel rolls. A moment later, a different material point is touching down, and that new point is the one at rest. The same idea gives a ruler construction. Here is a moving bar with the velocity direction already attached at A and B. Each point circles C, so draw a perpendicular to each velocity. The centre must lie on both lines. The two lines meet here, at C. From that one point, every speed is omega times the distance out to the material point. There is one limiting case. If the two velocities are equal and parallel, their perpendiculars are parallel too. They never meet, so C is at infinity and the body is purely translating."},{"title":"The Instantaneous Axis and Screw Motion","startSeconds":226.22579166666668,"narration":"Lift the body into three dimensions. This solid extends through all three coordinate directions, with its material points locked into one shape. Which of those points, if any, are momentarily at rest? Ask the rigid-body relation. Choose A on the body, draw its velocity, and keep the body's one angular velocity omega in view. We are hunting for a point P whose velocity is zero. The cross-product term must cancel v A exactly. It can do that only with the part square to omega, because every omega cross r is perpendicular to omega. Here the magenta arrow is that exact cancellation. For the moment v A is entirely square to omega. Solving the cancellation gives this particular displacement r zero. But r zero is only one answer. Add any multiple lambda omega and the cross product does not change. Watch lambda sweep through its values: every point it reaches is another solution, so the solutions fill a line. That tilted line is the instantaneous axis of rotation. Every green location on it has zero velocity at this instant. A point off the axis swings around it in a circle lying square to omega. Its speed is omega times its perpendicular distance from the axis. The same construction at B uses the same omega. Turn the view and the geometry separates cleanly: the red axis is one tilted line in space, and both green velocities stand square to it. Omega belongs to the whole body. A and B do not get different angular velocities. What changes from point to point is r, and therefore the cross-product contribution. Now remove the assumption we just used. If v A is not square to omega, split it into a perpendicular piece and a parallel piece. Let those two pieces land. Complete their parallelogram, and its green diagonal is the original velocity v A. The cross product can cancel the perpendicular piece, exactly as before. It can never touch the parallel piece, whose projection formula is this. So the axis still exists, but its points now slide along omega instead of resting. This yellow arrow is the velocity shared by every point on it. Take a point away from the axis. Its velocity is the sum of a turn around the axis and that same slide along it. Follow one material point. The combined motion traces this yellow helix, winding around the tilted axis while advancing along it. This is Chasles' theorem. At every instant, rigid-body motion is a screw: a rotation about an axis together with a translation along that axis. Watch the numbered point perform both parts. The slide per unit turn is the pitch. First limiting case: let the slide shrink to nothing. The advancing helix closes into a circle, the axis velocity vanishes, and the motion becomes pure rotation. Second limiting case: restore the slide and let omega shrink to nothing. The winding opens into a straight path, the red omega arrow disappears, and equal green velocities show pure translation. Put turn and slide back together. The tilted axis, the helix and the travelling point return as the one picture that contains the general case. Three ideas carry the lecture. Keep the moving screw beside them while we read the list. First, any two points share one omega, and their velocities differ by omega cross the vector between them. Second, plane motion has an instantaneous centre, but the material point occupying it changes as the body moves. Third, spatial motion has an instantaneous axis. Rotation about it plus translation along it is the general screw motion, every instant."}]}}