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Honors Physics 1 Class 06 Fall 2013 Newton’s Second Law as he wrote it Momentum and force Impulse Center of mass motion Center of mass 1 Newton’s Second Law Revisited d (ma ) to allow for the possibility dt that the mass could change (e.g.- be ejected). It turns out that the product ma is very useful in itself. It is called the momentum. We have previously only treated the motion of a point particle or masses with simple interactions. Momentum allows us to more easily calculate the behavior of multi-particle systems. Consider the motion of a pair of particles that only interact with each other (no external force on the "system".) dp dp f1 1 ; f 2 2 dt dt Since Newton's third law states that the force particle 1 exerts on particle 2 is equal and opposite to the force exerted by 2 on 1, we can write: d f1 f 2 p1 p2 0. This says that the momentum of the system is dt a constant of the motion, unless an external force acts on it. Newton actually wrote his second law as F 2 External force and momentum If there are both external and internal forces on a system of particles, Newtons 2nd law is: Fext fint Fext dpi . But the sum of all internal forces is zero, so i dt dP where P pi . dt i This equation looks just like our old single particle one. Extending the similarity: Fext MR, where M is the total mass and R can be determined by comparison the use of P. mi ri dP d MR pi mi ri which is true if R i . dt dt i M i R is called the Center of Mass of the system. Note that the equations we have deduced only apply to the translational motion of the center of mass (and not rotation.) 3 Momentum Because it is conserved in the absence of external forces momentum has great use in the analysis of collisions. Consider two bodies once again interacting in isolation. p1 (t1 ) p2 (t1 ) p1 (t2 ) p2 (t2 ) Example: A student sits on a stationary low-friction cart. The mass of the cart and student together is M=50 kg. The student throws an iron ball of mass 7 kg at a horizontal speed of 10 m/s in the +x direction. What is the motion of the student and cart after the throw? (Ignore friction.) Initial momentum of ball, cart, student = 0=psc pb therefore psc pb . pb mb vb 70 (kg m/s)iˆ 70 vsc iˆ iˆ1.4 m/s. 50 4 Discussion Two pucks collide elastically*. Using only conservation of momentum, what can you tell about the final velocities? *You know m1, m2, v1i, v2i. 5 Elastic and inelastic collisions Momentum is still conserved, independent of the type of collision. Elastic collision: Just use conservation of momentum. For a completely specified initial condition (m1, m2, v1, v2) , we have one equation, two unknowns. From what me know so far, we can deduce the relationship between the velocities, but not their magnitude. Completely Inelastic: The objects move off together after colliding. This means that we have one equation and one unknown. Easy peasy. 6 Different reference frames Look at the previous example from a reference frame that is initially moving wrt the cart. Momentum still conserved, but it is not the same. Change in speed is still the same. Initial: vsci vsc v fr v fr ; psci msc v fr Initial: vbi v fr ; pbi mb v fr Final: vbf 10 v fr ; Pi Pf ; msc mb v f (10 - v fr )mb vscf v f msc Same as problem on previous page from here. 7 Motion in the center of mass frame The laboratory is actually an arbitrary reference frame. We can choose any other inertial frame as well. Sometimes the center of mass is useful. CoM frame is the one in which the initial momentum of the system is zero. Example: One dimensional collision m1: Moves at velocity v1i in lab frame. m2 : Moves at velocity v2i in lab frame. Primed coordinates are the cm frame. v '1i v1i vcm ; v '2i v2i vcm P 'i m1v '1i m2 v '2i m1 v1i vcm m2 v2i vcm m1v1i m2 v2i vcm m1 m2 The center of mass system is the one where P ' 0 m1v1i m2 v2i vcm m1 m2 vcm m1v1i m2 v2i m1 m2 The final momenta in the cm system are easy to deduce. 1) Elastic collision: Particles bounce back out with their momenta reversed. 2) Completely Inelastic: Particle stick together and have zero motion. 3) Explosive: Particles leave with equal and opposite momenta from each other. 8 Calculating the Center of Mass The center of mass is found by integration over the volume. 1 1 1 xcom xdM x ; ycom ydM y ; zcom zdM z m m m You can frequently compute the center of gravity/mass using simple symmetries. 9 Example: Calculating the COM Let's try the calculation for a T-square that is 40 cm long with a T top that is 30 cm across, and where all of the parts are 3 cm wide. Take the thickness to be 2 mm. 1) Make a sketch. 2) Choose a set of axes. 3) Set up the integrals by breaking them up into simple sections. 4) Compute. 10 Calculating the COM y The T-square is symmetric about x x=0, so the xcom is at x=0. It is also symmetric about z=0. ( zcom 0) We have to actually do the y-integral. ydzdxdy 1 ycom ydzdxdy V dzdxdy The thickness is constant over the whole area, so the z integral is easy. 0 1.5 cm 1mm 3 15 cm 1mm 0 15 cm 1mm z Vycom y ( ( dz )dx)dy y ( ( dz )dx)dy 37 top view Vycom 1.5 cm 1mm 0 1.5 cm 3 15 cm 2mm y ( dx)dy y ( dx )dy 0 15 cm 37 1.5cm 11 T-square COM continued Vycom Vycom 3 0 0.2cm 3cm ydy 30cm ydy 0 37 2 3 2 0 y y 0.2cm 3cm 30cm 2 0 2 37 Vycom 0.3 0 37 2 3 32 0 0.3 1369 27 410.7 27 383.7 3 15 0 1.5 V dydxdz dydxdz 0.2 dydx dydx 0 15 0 15 0.1 37 1.5 0.1 37 1.5 3 0 V 0.2 3 dy 30 dy .6(37) 6(3) 22.2 18 40.2 0 37 Vycom 383.7 9.5cm ycom 40.2 V 0 1.5 0.1 3 15 0.1 12 T-square COM by serial logic Identify pieces of the object for which you can guess the COG. (Rectangles, circles…) Replace the piece with that mass at the COG. Find the balance point of the masses. COG of top piece is at (0,1.5,0) Mass(top)=Density*Vtop=D*18 COG of bottom piece is at (0,-18.5,0) Mass(bottom)=D*Vbottom= D*22.2 13 T-square COM by serial logic Find the balance point: y y DV y y DV y y V y y V 1.5 y 18 y 18.5 22.2 cogtop cog cogtop cog cog top top cog cog cogbottom cogbottom bottom bottom cog 27 18 ycog 22.2 ycog 410.7 22.2 18 ycog 40.2 ycog 383.7 ycog 9.5cm ycog ycogtop ycogbottom 14 Center of Mass Integration KK p 145 note 3.1 KK p119 nonuniform rod KK p120 triangular sheet 15 Center of mass integration: Nonuniform rod Take the mass per unit length to be: 0 dM ( x)dx L L x . L 2L L 0 0 x 0 L2 0 L x M dM dx 0 dx xdx L L L 20 L 2 2 0 0 0 1 R M 1 xiˆ yjˆ zkˆ dM M 1 xiˆ 0 ˆj 0kˆ dx M L 0 x x L dx 0 16 Center of Mass Integration We calculate the center of mass for a uniform triangular plate with edges at x 0, y 0, and y a 2 x 1 1 1 R rdm ( x , y ) r ( x , y ) dxdy ( x, y ) r ( x, y ) dxdy M M M M 4M Since the density is constant, = we can pull it out of the integral. 2 A a R xiˆ yjˆ dxdy M Rx a /2 a 2 x xiˆ dxdy M M 0 a /2 ax 2 x3 2 M 2 3 0 0 xdxdy M a /2 x a 2 x dx 0 3a3 2a 3 4 a 3 a 2 M 24 24 a 24 6 17