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4. Equation of Motion Using Angular Momentum

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4. Equation of Motion Using Angular Momentum Chapter 1 Introduction Section 1 Newton’s Laws dp Newton’s three laws: = F dt For a point particle: m m · First law (Law of inertia)[C]: It is always possible to find a p = v = r coordinate system with respect to which isolated bodies move For a rigid body: linear momentum of the CM. uniformly. This coordinate system is called an inertial reference 2 frame. d r = F dt2 The usual statement “A body remains in its state of rest or in Third law (C): If a given body A exerts a force on another body B, uniform linear motion as long as no external force acts on it to then body B also exerts a force on body A with equal magnitude change” is not suitable. In an accelerating frame, a force-free but opposite in direction. body will appear to accelerate. Not always valid. If the Sun moved by a distance, then the action Inertial reference frame: In which the law of inertial is valid. Any and reaction with Earth will not take on the new values until the reference frames moving with a constant velocity wrt stationary signal reaches the Earth. This is because the Newtonian frame is a inertial reference frame. dynamics assumes instantaneous propagation of signal (action at a distance). Examples: free space, a freely falling elevator. Main points: Second order differential equation. Second law (C): The change in the linear momentum of a body is proportional to the external force acting on the body and is in Other assumptions: the direction of the force. 2 Absolute space: Motion takes pace in the background of (1) Solve for the position and velocity of a damped linear oscillator absolute space. The space is same for all observers. whose mass m=1, the viscous coefficient is γ. The initial condition is [x(0),v(0)]. Consequence: The length of a rod is the same for all observers. (2) Do the same for a forced undamped oscillator with initial Absolute Time: Evolution of physical system in “time”. Time condition is [x(0),v(0)]. Solve for both cases: ω0 ≠ ωf and ω0 = ωf moves uniformly with equal rate for all observers. (3) A spherical ball of mass m falls under gravity in a viscous Consequence: Time interval between two events is the same fluid. Find the position and velocity of the ball as a function of for all observers. time. Assume that the mass starts at rest from a height h Mass conservation above the ground. Apply the above solution to a raindrop whose radius is 1 mm. Assume the dynamic viscosity of air to Infinite speed be 10−5 kg/(m s). Examples: Oscillator: Equation ·· · 2 m x + 2γx + ω0 x = F0 cos ωf t (1) Linear frictionless oscillator with the initial condition is v(0) [x(0),v(0)]: x(t) = x(0)cos ωt + sin ωt ω (2) Solve x·· = x: x(t) = x(0)cosh t + v(0)sinh t Exercises: 3 Section 2 Energy Work energy theorem: Conservative force fields: Fields for which F ⋅ dr = 0 for any The work done on a particle by a force F as it moves from point ∮ Q A to point B: closed path. For such fields, F ⋅ dr is only function of the end ∫P B W = F ⋅ dr points, but remain independent of paths. Therefore, we choose AB ∫ A P as the reference point and define potential energy U(r) as which can be rewritten as r U(r) = − F(r′)! ⋅ dr′.! ∫ B P B dv dr 1 [1] W = m ⋅ dt = mv2 = T − T AB ∫ dt dt [ ] B A A 2 A Therefore, B B A where TA, TB are the kinetic energy of the particle at the points A [2] F ⋅ dr = F ⋅ dr − F ⋅ dr = − UB + UA and B respectively. ∫A ∫P ∫P The above statement is work energy theorem: The difference in Eqs. [1,2] yield the kinetic energy of a particle between two points is equal to the net work done on the particle by the external forces during the TA + UA = TB + UB, transit. The sum of kinetic and potential energies is the total energy, and Conservative force fields and Potentials: the above statement is the conservation of total energy of a mechanical systems that are conservative. 4 Example: these charges. (a) Find the position(s) of equilibrium point(s). (b) Compute the potential near the equilibrium point. (c) What (1) Spring force F k x. Therefore = − is the nature of the equilibrium point? Describe the motion of x 1 1 U(x) = − (−k x)dx = k x2 − k x 2 the test particle near this point. (d) Does your analysis depend ∫ 2 2 0 x0 on the sign of the test charge? Hence, the potential energy is U(x) = k x2 /2 apart from a constant, (3) What is the potential of an electric dipole? Compute the and the total energy electric field induced by it. Sketch the potential. 1 1 (4) Philip Morse modelled the potential energy of a diatomic E = m x· 2 + k x2 2 2 molecule using a potential energy function: U r D a r r 2 where r is the distance between (2) For gravitational force F = − GMm /r2 r̂ , ( ) = e{1 − exp[− ( − e)]} the atoms of the dipolar molecule re is the equilibrium bon B B 1 1 distance and De is the depth of the potential well. (a) Sketch U(B) − U(A) = − F ⋅ dr = − Frdr = − GMm − ∫A ∫A ( rB rA ) the potential. (b) Find the minima of the potential. (c) Expand the potential near its minimum and compute the frequency of By choosing r , we deduce A = ∞ oscillation. GMm U(r) = − r Exercises: (1) A charged particle +Q is fixed at the origin. A test charged particle +q of mass m is fired head-on towards the charged particle +Q with velocity v∞ from ∞. Compute the shortest distance of approach between the particles. (2) Four immovable charges of magnitude +Q are placed at the vertices of a square. A test charge +q is moving in the field of 5 6 Section 3 System of Particles and Conservation Laws 1. Linear Momentum fa = fa,ext + fa,int = fa,ext + ∑ fa,b b -th particle. Centre of mass: where fa,b is the force on the a-th particle due to the b m ∑ ara MaCM = fa = fa,ext + fa,b RCM = ∑ ∑ ∑ ∑ ∑ ma a a a b a: Particle label From Newton’s third law: CM reference frames: Frame in which RCM = 0. fa,b + fb,a = 0 Linear momentum of a system: Therefore, ! d P P = mava = ma(VCM + va′) = M VCM = PCM = f = F . ∑ ∑ dt ∑ a,ext ext a Equation of motion: If Fext = 0, then the total linear momentum of the system P is a · d VCM P = M = m v· = f constant. This is the statement of the conservation of linear dt ∑ a a ∑ a a momentum. fa is the net force on the a-th particle. Example: Two masses (m) connected by a spring. Example Ch 13 7 If the torque, N, on the particle is zero, then the angular momentum L of the particle is a constant. This law is called conservation of angular momentum. 3. Angular Momentum of a System of 2. Angular Momentum of a Single Particle Particles y The Angular momentum of a system of particles is a sum of V angular momentum of individual particles. The angular P .m momentum of such system about a reference point is r L = ∑ ra × pa, . a O where denotes the position vector of the ath particle from the x ra reference point, and pa its linear momentum. Using the CM ! ! We define the angular momentum of the particle about a coordinates, r = RCM + ra′ and v = VCM + va′: reference point O, where OP = r, as L = ∑ ra × pa a L = r × p, = m (R + r′!) × (V + v′!) · ∑ a CM a CM a where p = mr is the linear momentum of the particle measured in a the reference frame. The above formulas is valid in any reference = × + ′! × ′! frame, inertial or noninertial. Now. RCM PCM ∑ ra pa Therefore, d L dp dr dp = r × + × p = r × = r × F = N dt dt dt dt L = RCM × PCM + LCM = Lorbit + LCM, where where N is the torque on the particle about the reference point O. 8 ! ! LCM = ∑ ra′ × pa′ is the angular momentum of the body about the CM, The time derivative of LP is d LP ·· = m (r − R ) × (··r − R ). dt ∑ a a P a P 4. Equation of Motion Using Angular a Momentum ·· Using mara = fa is the net force acting on the ath particle, we obtain We assume that xy coordinate system of the figure is an inertial d LP ·· reference frame, and the reference point O belongs to it. The = (ra − RP) × (fa − maRP). dt ∑ a angular momentum of the body about O is ·· The term (ra − RP) × (−maRP) can interpreted as the torque on LO = mara × pa. ∑ ∑ a a ·· the rigid body due to the pseudo force −maRP. The force fa is a We also compute angular momentum LP about another point P sum of external and internal forces, i.e., shown in the following figure: fa = fa,ext + ∑ fa,b m ! ! m · · , b LP = ∑ ara′ × pa′ = ∑ a(ra − RP) × (ra − RP) a a where is the external force on the th particle, while is the fa,ext a fa,b internal force on the ath particle due to the bth particle as shown y in Fig. 15.4. Therefore, d r'a LP ·· 1 r = (r − R ) × (f − m R ) + (r′! × f + r′! × f ) a dt ∑ a P a,ext a P ∑ ∑ a a,b b b,a P a 2 a b RP O x 9 where ! .
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