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,