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Outline CONSERVATION OF MOMENTUM Ozkaya and Nordin p. 317-335 Newton’s second law of motion Linear motion: If a non-zero force F acts on a rigid body of mass m, the body’s centre of gravity will have an acceleration aCG, according to the law F=maCG. Angular motion: If there exists a non-zero moment MCG about a body’s centre of gravity, the body will have an angular acceleration !, according to the law MCG=ICG!, where ICG is the moment of inertia of the body about its centre of gravity. ma ! Fx = m(aCG ) x ! Fy = m(aCG ) y ! MCG = ICG" F2 CG F1 • Newton’s equations of motion (review) • impulse-momentum • conservation of linear momentum • conservation of angular momentum • moment of inertia Newton’s second law is the law of conservation of momentum conservation of linear and angular momentum d (mvx ) dt d Fy = ( mv y ) dt Fx = MCG = law of impulse-momentum t2 ( !t t !t Fy dt = m(vy 1 Fx dt = m vx 1 " vx 2 2 ICG! F3 F4 d (I ! ) dt CG 1 1 " vy 2 t2 ) !t ) 1 Mdt = ICG (# 1 " # 2 ) Ground reaction forces mirror COG acceleration vertical GRF >BW: upward acceleration or downward deceleration (of magnitude ((Fy/(body mass)-g)) vertical GRF >BW: downward acceleration or upward deceleration 3 COG vertical position (m) Walk Run Vertical Force Fy (BW) Ground reaction forces (measured from a force plate) mirror acceleration of the whole-body centre of gravity. Ground reaction forces during walking COG vertical acceleration (m/s2) 2 RHS RHS 0 0.0 1 upward acceleration 2 upward deceleration 200 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Time (seconds) Activity Walking Running VB Spike Landing Somersault Landing Peak Vertical GRF 1.1 BW 2.8 BW 4.8 BW 13.6 BW Ground reaction forces during running high-frequency “heel strike transient” (occurs during running but not walking) LHS RTO 1 3 downward acceleration 1 Foot Contact 150 Forces (%Body Weight) 100 Vertical Med (+) - Lat (-) 50 Post (+) - Ant (-) 2 4 3 4 downward deceleration 5 forward braking 6 forward propulsion 5 0 6 -50 0 25 50 75 100 % Gait Cycle Law of impulse-momentum Impulse applied to the system between instants 1 and 2 t2 = ( t !t Fy dt = m(vy !t 1 Change in momentum of the system Fx dt = m vx 1 " vx 2 2 1 t2 !t 1 1 " vy 2 ) ) Mdt = ICG (# 1 " # 2 ) Computing the impulse (a) Computing the impulse (b) force (N) force (N) 3 3 2 2 1 1 1 2 3 4 5 6 7 time (s) impulse = 1 2 3 4 5 6 7 time (s) impulse = Ground reaction forces during vertical jumping Computing the impulse (c) vert. GRF 800 N time Each box: 100 N represents (100 N)(0.05 s) = 5 Ns 0.05 s “squat” jump impulse = “countermovement” jump Conservation of linear momentum If the net force applied to a system is zero, then the second law states : d d (mvx ) = 0 (mvy ) = 0 dt dt and the momentum in the x and y directions is constant (conserved) : (mvx ) = constant = px (mvy ) = constant = py Newton says: if there’s no force, there’s no change in momentum. Therefore, if m increases, v must decrease. This is the law of conservation of linear momentum : m1v x1 = m 2v x2 Linear momentum has units of [kg][m][s-1] m1v y1 = m 2v y2 Conservation of angular momentum If the net moment applied to a system is zero, then the second law states d (I ! ) = 0 dt CG and the angular momentum is constant (conserved) : ( ICG! ) = constant = H This is the law of conservation of angular momentum : ( ICG )1! 1 = ( ICG ) 2! 2 If I increases, ! must decrease, and vice - versa. v -v 2v m m m 0 m v v=0 m m m v -v/2 m 2m v=0 3m Moment of Inertia The moment of inertia I of an object about a specific axis is 2 I = ! mi ( ri ) = " r 2 dm i where r is the distance from the mass mi (or mass element dm) to the axis of interest. For the Y object shown at left : 2 I X = ! mi ( yi ) = " y 2dm i 2 IY = ! mi ( xi ) = " x 2 dm i m 3v/2 0 m 2m v/2 3m The moment of inertia reflects that a body has an inertia for turning that depends not only on the masses, but also on how far away they are from the axis of rotation. X Clearly, Ix > Iy in this example Moment of inertia has units of [kg][m2] Example: computing moment of inertia Calculate the moment of inertia of the object below (a) about the axis y1 (which passes through the CG of the system), and (b) about the y2 axis. y2 y1 20 kg 10 kg 0.5 m "miri2 1.0 m Iy1 = = (20 kg)(0.5 = 15.0 kg·m 2 m)2 + (10 kg)(1.0 m)2 Iy2 = "miri2 = (20 kg)(1.5 m)2 + (10 kg)(0 m)2 = 45.0 kg·m 2 Example: Parallel Axis Theorem Same as last problem. However, calculate the moment of inertia about the y2 axis using the parallel axis theorem. y2 y1 20 kg 10 kg 0.5 m 1.0 m Iy1 = ICG = "miri2 = (20 kg)(0.5 m)2 + (10 kg)(1.0 m)2 = 15.0 kg·m2 Iy2 = I = ICG + mTOTAL(rCG)2 = 15.0 kg·m2 + (30 kg)(1.0 m)2 = 45.0 kg·m 2 The Parallel Axis Theorem If the moment of inertia about an axis passing through the center of gravity is known, the moment of inertia about a second, parallel axis is given by: I = ICG + mTOTAL(rCG)2 where ICG = MOI about axis passing through the CG mTOTAL = total mass of the object rCG = perpendicular distance between axes Conservation of angular momentum during free-flight During free-flight, the net moment about the body’s COG is zero. Therefore, while ICG and " vary considerably, the product H= ICG " must remain constant. Conservation of angular momentum during a long jump avoiding forward rotation of the trunk after take-off late stage preparation for landing Conservation of angular momentum during a cat flip The cat appears to defy the law of conservation of angular momentum, by rotating forward and hind segments in different directions, and appropriately drawing the legs in an out to modify the MOI of each segment. GRF Angular momentum is conserved, but energy is not. A classic example of conservation of angular momentum is the figure skater pulling her arms in (and decreasing her ICG) to increase her ". Angular momentum is conserved, but energy increases. Where does the energy come from? (HINT: think about the centrifugal force acting to pull her hands outward) Momentum is a vector, and must be conserved about a given axis Another classic example of conservation of angular momentum. When the spinning wheel is moved so its axis of rotation is vertical, the frictionless turntable spins in a direction opposite to the wheel (conserving angular momentum about the vertical axis). Review Questions Under what conditions is a body’s linear momentum constant? Angular momentum? ! Can you compute a linear impulse? An angular impulse? ! When angular momentum is constant, what is the effect of changing moment of inertia? ! What influences the moment of inertia of the human body? ! Can you compute moment of inertia? ! Can you apply the parallel axis theorem? ! Can you apply the principles of conservation of linear momentum/ conservation of angular momentum? ! What is the “cat twist” technique? ! How does a (long jumper, gymnast, volleyball player) control movement while in flight? ! The gymnast shown in the figure at right loses balance while performing on the beam, and begins to rotate counterclockwise (CCW). (a) Mark on the diagram (with arrows) the direction she should move her arms to help her regain balance. (b) Why does this work? Explain the physics underlying this phenomenon. Example problem: cycling around a turn Example problem: moment of inertia Show that a narrow rod with uniformly - distributed mass (often used to model body segments) has (a) a moment of inertia about its center of 1 gravity equal to I CG = ml 2 , 12 and (b) a moment of inertia 1 about its end equal to I0 = ml 2 . 3 Example problem: conservation of angular momentum 0 CG l Consider a cyclist of body mass m going around a turn of radius r, with constant angular velocity ". (a) Identify the magnitude and direction of the cyclist’s velocity and acceleration. (b) Mark on a free body diagram the centrifugal force applied to the cyclist’s centre of gravity, and the associated centripetal force applied to the tires of the bike. (c) Explain, using a free body diagram, why cyclists must lean into a corner to keep their balance. (d) Explain, again with a free body diagram, why one should be able to cycle faster around banked corners. Conservation of angular momentum during free-flight (cont) Estimate the change in angular velocity between (b) and (d) below, given the ICG values shown at left. ICG = 3.5 kg·m2 ICG = 6.5 kg·m2 ICG = 15.0 kg·m2 IHBAR = 85.0 kg·m2