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PHY 5200 Mechanical Phenomena Momentum and Angular Momentum PHY 5200 Mechanical Phenomena Newton’s Laws of Motion Claude A Pruneau Click to edit Master title style Physics and Astronomy Department Wayne State University Dec 2005. Claude Click A Pruneau to edit Master subtitle style Physics and Astronomy Wayne State University 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 1 Principle of Conservation of Momentum • We consider the total momentum of a system of N particles. – we label each particle with =1,…, N. N r r r r P p1 p2 L pN p 1 • Provided the internal forces obey Newton’s third law, one can write d r r P Fext dt • This implies Fext 0 P0 P constant If the net external force Fext acting on an N-particle system is zero, the r system’s total mechanical momentum P m v is constant. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 2 Example - Inelastic Collision of two bodies Question: Two bodies have masses m1 and m2, and velocities v1 and v2. The two bodies collide and lock together - and subsequently move as a single unit. Note: This is perfectly inelastic collision. Assuming external forces can be neglected during (and shortly after) the collision, find the velocity of the combined masses v. Before Collision After Collision 2 v2 2 1 5/22/2017 v1 1 Claude A Pruneau, PHY5200, Chap 3 v 3 Example - Inelastic Collision of two bodies (cont’d) Solution: Use the fact there are no external forces acting on the two bodies. That implies the total momentum of the system is conserved (I.e. constant). r r r r Pinitial p1 p2 m1v1 m2 v2 r r r Pfinal m1v m2 v m1 m2 v r Pfinal Pinitial r r r m1v1 m2 v2 m1 m2 v r r m1v1 m2 v2 v Weighted average of the initial velocities. m1 m2 Important Special Case: One of the two masses initially at rest: r m1v1 v m1 m2 5/22/2017 Direction is the same, but the speed smaller Claude A Pruneau, PHY5200, Chap 3 4 Motion of rockets • • • • • QuickTi me™ and a TIFF ( Uncompressed) decompressor are needed to see thi s pi ctur e. Rocket propulsion provides an excellent example of applicability of “momentum conservation”. Rocket motion is achieved from ejection of gas at high speed - no eternal force is required to push or pull the rocket. F=ma is not straightforwardly applicable given the mass is constantly changing because of fuel consumption and ejection. F=dp/dt is however applicable. Consider vex 5/22/2017 v v Speed of the rocket relative to the ground. vex Speed of the exhaust gas relative to the rocket. Claude A Pruneau, PHY5200, Chap 3 5 Motion of rockets (cont’d) • QuickTi me™ and a TIFF ( Uncompressed) decompressor are needed to see thi s pi ctur e. At time t, the rocket has a momentum P(t). P(t) m(t)v • An instant dt later, i.e. at time t+dt, the mass of the rocket is reduced m(t t) m(t) dm • Its speed at that time is v(t t) v(t) dv • The momentum of the rocket is • • The fuel ejected during the time interval dt is -dm The velocity of the fuel relative to the ground is P(t t) m(t) dm v(t) dv v(t) vext • The momentum of the fuel ejected during the time interval v(t) v(t) vext vext dm v(t) vext 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 6 Motion of rockets (cont’d) • In the presence of gravity one would have P(t t) P(t) FG t • • For simplicity, assume no gravitational force, FG = 0. Apply momentum conservation 0 P(t t) P(t) 0 P(t t) P(t) m(t)v(t) m(t) dm v(t) dv dm v vext • Distribute products, drop (t) mv mv mdv vdm dmdv vdm vext dm • The above expression is evaluated in the limit t0, dmdv is thus negligible. mv mv mdv vdm vdm vext dm • Simplify terms in “mv” and “vdm” 0 mdv vext dm 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 7 Motion of rockets (cont’d) • One has 0 mdv vext dm • Or mdv vext dm • Divide both sides by “dt” m • Or • Where 5/22/2017 dv dm vext dt dt r& mv vext m& vext m& has units of a force and is called “thrust” Claude A Pruneau, PHY5200, Chap 3 8 Rocket Motion - Solution 1 • The rocket motion is governed by m& v vext m& • Or equivalently dv dm m vext dt dt • Eliminate “dt” mdv vext dm • Separate the variables dm dv vex m 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 9 Rocket Motion - Solution 2 • We need to integrate dm dv vex m • • • • • The “initial” velocity is “vo”. The “final” velocity is “v”. The “initial” rocket mass will be labeled “mo”. The “final” is written “m”. The integral can thus be written v m dm v dv vex m m o o • • Where it is understood that both “v” and “m” are functions of time “t”. Integration yields: v vo vex ln m mo m 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 10 Rocket Motion - Solution 3 • Simplification yields v vo vex ln m ln mo • Or equivalently • Or more simply v vo vex ln m / mo v vo vex ln mo / m 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 11 Rocket Motion - Notes • The solution is v vo vex ln mo / m Notes: • • • • The mass mo, the initial mass of the rocket, includes fuel and payload. This result places a severe constraint on the maximum speed of a rocket. The ratio mo/m is largest when all the fuel is burned and “m” corresponds to the rocket+payload mass. If the initial mass is 90% fuel, then the ratio is “10”, since ln(10)=2.3, the gain in speed only amounts to 2.3 the exhaust velocity. – Rocket builders try to maximize the exhaust velocity, and/or design rockets with multiple stages. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 12 Saturn V Rockets QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. July 16th, 1969 launch oF Saturn V rocket carrying the crew of Apollo 11 to the Moon. The Saturn V rocket was the largest rocket ever used by NASA, and the only one able to lift the large masses needed to land astronauts on the moon and returning them safely. Saturn V rockets launched all of the Apollo moon missions, and several to Earth orbit as well. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 13 QuickTime™ and a TIFF (Uncompressed) decompressor are needed to see this picture. Size Height Diameter Mass Stages 111 m (364 ft) 10 m (33 ft) 2,900,000 kg (6,500,000 lb) 3 (2 for Skylab launch) Capacity Payload to LEO 75,000 kg (2-stage) Payload to the Moon 118,000 kg (3-stage) 47,000 kg First Stage - S-IC Quic kTime™ and a TIFF (Unc ompres sed) dec ompres sor are needed to see this pic ture. Engines Thrust Burn time Fuel 5 F-1 engines 33.4 MN (7,500,000 lbf) 150 s RP-1 and liquid oxygen Second Stage - S-II Engines Thrust Burn time Fuel 5 J-2 engines 5 MN (1,000,000 lbf) 360 s Liquid hydrogen and liquid oxygen Third Stage - S-IVB Engines Thrust Burn time Fuel 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 1 J-2 engine 1 MN (225,000 lbf) 165 + 335 s (2 burns) Liquid hydrogen and liquid oxygen 14 Center of Mass • Many ideas seen so far can be rephrased in terms of the notion of center of mass. • Consider a group of N particles, labeled =1, …, N, with masses m and positions r measured relative to some arbitrary origin O. 1 2 r1 r2 r3 rN 3 N 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 15 C. M. of a continuous system • • • The notion of C. M. was 1st defined for a system of N particles. In practice, one often deals with extended and continuous objects. So we must consider the notion (extension) of CM for continuous objects. • The discrete sum over all particles, must be replaced by a continuous sum over all infinitesimal elements of mass, dm. N m i i 1 • • • • dm volume Where the integral must be carried over the entire volume I.e. to account for all mass elements. Here we face an integral in “dm” taken over a “volume”. It is therefore more practical to transform the integral into a volume integral. Introduce the mass density or density m v 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 16 C. M. of a continuous system • • The mass density may not be uniform across an object. – E.G. in the human body, the bones, muscles, and other internal organs vary in density. The notion of density must therefore be defined for infinitesimal mass or volume elements. dm r (r ) • • Note that the density is here noted as a function of the position. With this expression in hand, it is now possible to write the expression for the CM of continuous system. rCM • • dV 1 M 1 r rdm M volume r dm r dV dV volume Where the last expression indeed corresponds to a volume integral. Insert the definition of density in the above expression. rCM 5/22/2017 1 M r r r (r )dV volume Claude A Pruneau, PHY5200, Chap 3 17 C. M. of a continuous system (cont’d) • • The expression for the center of mass is a vector equation. It can be decomposed into three Cartesian components. rCM • • 1 r X x ( r )dV cm M volume 1 1 r r r r ( r )dV Y y ( r )dV cm M volume M volume 1 r Z z ( r )dV cm M volume Also note the volume element dV may be written as a product dV=dxdydz. The above integrals are thus of the form Xcm 5/22/2017 1 M r x (r )dxdydz volume Claude A Pruneau, PHY5200, Chap 3 18 C.M. Continuous System - Example 1 • • Calculate the position of the C.M. of a rectangular box of sides a, b, and c and uniform density . Let’s perform the calculation in a reference frame oriented along the sides of the box as follows: z c a x 5/22/2017 y b Claude A Pruneau, PHY5200, Chap 3 19 Example 1 - cont’d • • • The calculation proceeds similarly for all three directions x, y, and z. We show the calculation along “x” only. The definition of c.m. along x is Xcm • 1 M volume M xdxdydz volume The volume integral is calculated on a rectangular volume and may thus be written: Xcm • x dxdydz The density is constant and can therefore be factorized out of the integration. Xcm • M c b a 0 0 0 dz dy xdx This yields Xcm 5/22/2017 a x2 a2 z 0 y 0 bc M 2 0 M 2 c b Claude A Pruneau, PHY5200, Chap 3 20 Example 1 - cont’d • We have Xcm • a2 M 2 bc The mass can also be written as the product of the density and the volume of the box. M V abc • The C.M. position (along x) is thus Xcm • a2 abc 2 bc a 2 By symmetry, the answer is similar along the y and z axes. The C.M. position of a rectangular box of sides a, b, and c is thus: a b c rcm , , 2 2 2 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 21 Example 2 C.M. of a solid cone • Let’s calculate the C.M. position of the solid cone illustrated below. z • By symmetry, the position of the c.m. along x and y is (0,0). • We need only calculate the position of the c.m. along “z”. R h R=Rz/h y O x 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 22 Example 2 - Cont’d • The position of the c.m. along z is given by Z cm • M zdxdydz volume Given the cylindrical symmetry, it is more convenient to carry the integration in cylindrical coordinates (,,z). Z cm • • M zd d dz volume The density is assumed constant and can thus be factored out. The integral may be written Z cm h Rz /h 0 0 zdz M 2 d d R h R=Rz/h y 0 x 5/22/2017 z Claude A Pruneau, PHY5200, Chap 3 O 23 Example - cont’d • • The integral is Z cm h Rz /h 0 0 zdz M 2 d d 0 Integration over yields a factor 2. Z cm 2 Z cm 2 Zcm 2 h Rz /h 0 0 zdz M h h 5/22/2017 Rz /h zdz 2 M 0 0 M 2 2 Rz / h z dz 2 0 R 3 z dz h M 0 2 Z cm d h 4 R2 h 2 R h h M 4 4M 2 Z cm Claude A Pruneau, PHY5200, Chap 3 24 • You can verify that the volume of the cone is 1 2 V R h 3 • z The mass can be evaluated, and the CM position is thus Zcm R h 2 4M 2 R h 2 4 R 2 3 3 h R2 h 4 h R=Rz/h y O x 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 25 Angular Momentum • The angular momentum is defined as r r l rp • angular momentum • Note that the vector position r is defined relative to a given origin O, the angular momentum l, is thus also defined relative to that same origin, I.e. its value depends on the choice of the origin. One should therefore refer to l as the angular momentum relative to O. • Now consider the time rate of change of the angular momentum. • The derivative of a product is obtained with the product rule. r dl r& d r r l r p dt dt 5/22/2017 r r dr r r dp r& r r r& l pr r pr p dt dt Claude A Pruneau, PHY5200, Chap 3 26 • The time rate of change if l is r p mv r r r r l r& p r p& • Remember the expression of the force r r F p& • Also remember the momentum is parallel to the velocity. r r r l r& p r F r r l rp O 0 • We get an expression for the net torque about O. r r l rF • Note many textbooks use the letter to denote the torque, but here we will use the same notation as Taylor. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 27 Torque • • In many problems, one can choose the origin such that the net Torque is null. In such cases, that implies the angular momentum about that origin is a constant of motion. Example of the motion of planets around the Sun. • Clearly, r O 5/22/2017 GmMr̂ F r2 r& r l rF0 Angular momentum is constant r and p remain in the same plane. The motion of the planet (orbit) is confined to a plane. The motion is reduced to two dimensions… Claude A Pruneau, PHY5200, Chap 3 28 Kepler’s Second Law • • One of the greatest achievements of Newton lies in that he was able to explain the Kepler’s 2nd law as a simple consequence of conservation of angular momentum (Principia 1687). We will discuss Kepler’s first and third laws later, here we focus on the 2nd law. Kepler’s 2nd Law As each planet moves around the sun, a line drawn from the planet to the sun sweeps out equal areas in equal times. Q P dA O dA P’ 5/22/2017 Q’ r r dr vdt Claude A Pruneau, PHY5200, Chap 3 29 Q P dA • Calculate the area of the triangles O 1 r r dA r vdt 2 • dA P’ Q’ r r dr vdt Replace v by p/m, divide by dt. r dA 1 r p 1 r r l dt 2 m 2m • Since the planet angular momentum is conserved for a central force, then dA/dt is constant, hence we demonstrated Kepler’s 2nd law. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 30 Angular Momentum of Several Particles • For a system of N particles, =1, 2, …, N, each with angular momentum r r l r p • • All measured relative to the same origin O. The total angular momentum is defined N r r r L l r p N 1 • 1 Let’s now consider the torque r& N r r L l r F N 1 • 1 The rate of change of the angular momentum is simply the net torque. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 31 Net Torque • The net force acting on each particle can be written r r ext F F F • Remember action=reaction r F F • The angular momentum can thus be written r r ext r r L r F r F • Focus on the 1st term, it can be re-written r r r r r r F r F r F 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 32 r r r r r r F r F r F • Again invoking action = reaction r r r r r r F r F r F • Or equivalently r r r r r F r r F • Note that by construction, the above vector product is null because the two vectors are parallel to one another. r r r F r r O 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 33 • So, we conclude that only the external forces have relevant contributions to the net torque. r r ext r r L r F r F 0 r ext r L r F ext The time rate of change of the angular momentum is determined by the net torque. If the net external torque is null, the angular momentum of the system is constant. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 34 Moment of Inertia • Calculations of the angular momentum may often be simplified and formulated in terms of angular variables. • For example: Angular velocity of rotation Lz I Moment of Inertia Uniform disk (mass M, radius R) I 1 MR 2 2 Solid Sphere (mass M, radius R) I 2 MR 2 5 N In general I m 2 1 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 35 Net torque & C.M. • We found r ext r L r F ext • Where both L and G are measured relative to some origin in some reference frame. • We will state without demonstration that this result also applies for non inertial frames provided the origin is chosen to be the C.M.. 5/22/2017 Claude A Pruneau, PHY5200, Chap 3 36