Prof. O. B. Wright, Autumn 007 Mechanics Lecture Newton s Laws of Motion, Energy and Oscillations Reference frames e.g. displaced frame x =x+a y =y x =z t =t e.g. moving frame (t=time) x =x+vt y =y x =z t =t Inertial frame: A frame of reference in which Newton s laws of motion are obeyed. Example of a non-inertial frame is an accelerating frame or rotating frame. Newton s laws of motion These laws apply to point masses. The frame of reference is an inertial frame. Law no. : In the absence of forces on it, a particle moves with constant velocity dr v constant when F=0. (r is the displacement vector) Or r vwhen F=0. Law no. : For a particle of mass m, the total force on it F (called the net force) is F mr or F mv, where v r.
Prof. O. B. Wright, Autumn 007 In terms of momentum p mv, F p Total force = rate of change of momentum This law does not tell us where the force comes from. It just tells us that if a particle is seen to accelerate with acceleration equal to mr. The components of the force are F mx, F my, F mz x y z r, then we know that the total force acting on it is What is missing is the equations for the origin of the force. To complete the theory we must go outside of Newton s laws. For example we must write down the law or gravity or the laws of electromagnetism. Law no. 3: Whenever two particles interact, force F on particle (due to particle ) is equal and opposite to the force F on particle (due to particle ) at that instant in time. F=- F Simplest case is when the two forces central forces point towards or away from each other: e.g. F Gm m r / for gravitation But it is also possible to have this type of force in the 3rd law: (for example between two electric dipoles or±charge pairs) non-central forces Example of a system that can break the 3rd law! (for example between moving charges) Why can it be broken? ) Because in some cases the force depends on the history of the particle (where it has
Prof. O. B. Wright, Autumn 007 been before). (Effects cannot travel faster than light so forces are not really instantaneous.) ) Because electric and magnetic fields between the particles can carry momentum. Something to think about. How can masses or charges communicate their force across empty space? Conservation of momentum for two particles From the 3rd law: F+F=0 = total force acting on the system of two particles From the nd law: total force d F p dp dp d ( p p) 0 p p constant This is the law of conservation of momentum for the system of two particles. Introduction to mechanical energy Work W Definition: the work done by a force F: W F or, in 3D, Fx W F dr Fy dy Fx Fydy Fz dz F dz z Example: slowly lift a mass m a vertical distance x with a vertical external force of magnitude Fe. What is the work done by the external force Fe? 3
Prof. O. B. Wright, Autumn 007 Since the lifting is very slow, Newton s third law applies during lifting so Fe=-F, where F=-mg is the magnitude of the gravitational force (distances and forces are measured positive upwards). x x x e e 0 0 0 W F dr F F mg mgx, where g is the acceleration due to gravity (or the force per unit mass due to gravity). Potential energy V Definition: potential energy change is the work done by an external force during a slow change, as in the above example: V V V W Fe. dv For a small change dv Fe Fe. But Fe=-F for a slow change, so dv F. In the above example F=-mg so that V=mgx+const. Example: Consider stretching a spring that obeys Hooke s law F=- x, where F is the pulling force of the spring, x is the stretching displacement, and is the spring constant. Find the work done by the external force Fe, and check that F=-dV/. x e 0 W F F x x 4
Prof. O. B. Wright, Autumn 007 The work is only done by the force Fe on the right hand side of the spring because the oppositely directed left hand side force Fe does not move. To find V we can use V W Fe x. Therefore dv So F x as required. V x const. Kinetic energy T Definition T m v mv. v Example: throw a ball of mass m upwards with velocity v. Find the height h reached. Method : Use Newton s nd law: mx F mg x g Integrate: x gt v (used initial condition that x v at t=0) Integrate again: (used initial condition that x=0 at t=0) x gt vt v v x 0 t x h g g Method : Use conservation of energy: initial kinetic energy=final potential energy, so mv mgh. This gives the same result for h. 5
Prof. O. B. Wright, Autumn 007 Simple harmonic motion (SHM) Consider the uniform motion of a particle in a circle of radius a. Fig. : The displacement of the particle can be represented by a vector a rotating at angular velocity. Fig. : At the same instant the velocity vector is shown. It is normal (perpendicular) to r. This also rotates at, and has magnitude (absolute value) a. To see why consider r at times t and t+ t. = - t The velocity vector r is defined by where ê dr r( t t) r( t) d r Lim a eˆ a e ˆ, t 0 t is the unit vector in the direction of r (perpendicular to r) and we have used r a (because is a very small angle). Fig. 3: The acceleration displacement. This can be seen using r has magnitude a, and is opposite in direction to the r =dv/ as we did for v=dr/. The components of these vectors along the horizontal axis oscillate in SHM. The diagrams help you to see the relation between displacement, velocity and acceleration in SHM. If the position vector r in Fig. makes an angle with the horizontal axis when t=0, at any other time t it makes an angle - t, and the horizontal component of the displacement is given by 6
Prof. O. B. Wright, Autumn 007 x=acos( - t)=acos( t- ) From Figs. and 3 (or by direct differentiation) we see that the velocity x and acceleration x have the form: x a cos( t / ) a sin( t ) x a t a t cos( ) cos( ) a is the Amplitude of oscillation t- is the Phase at time t is the Phase Constant is the Angular Frequency The Period T is the time required for t to change by radians, so T = /. If T is measured in seconds, /T is the frequency measured in cycles per second or Hertz (Hz); is measured in radians per second (s - ). The phase is dimensionless. The amplitude of the oscillations of velocity is a and the amplitude of the acceleration is a. For SHM, the acceleration is always times the displacement x and oppositely directed. A mass m can therefore be maintained in SHM if acted on by a force m times the acceleration and directed towards the origin (opposite to the displacement). This is a Hooke s Law force. Differential equation of SHM x x 0 () 7
Prof. O. B. Wright, Autumn 007 A mass acted on by a Hooke s law force - x obeys the equation of motion mx x. () Therefore by comparison of Eqs. () and (), / m. Formal solution of differential equation: write x dv dv dv x v v, and Eq. () becomes v, so that dv v x 0. Rearrange and integrate: vdv x. So v ( a x ), where a is a constant of integration. (3) Hence v a x or. a x Put x=a cos, so that =-a sin d, and asin d ( t t0), where t0 is an integration constant. asin Hence x=acos( - t), where = t0. 8
Prof. O. B. Wright, Autumn 007 Energy in SHM Inserting /m for in Eq. (3) we obtain mv x a =constant. This is an expression of energy conservation, because and (/ ) x (/ )mv is the kinetic energy is the potential energy. It is possible to use this law of energy conservation to determine vibration angular frequency. 9