Review of Newtonian Physics1. Linear MomentumNewtonian physics describes the motion of point-like masses via Newton’s Second Law of Motion,F→ = m dv→dt ≡ dp→dtwhere F→ is the force acting on an object of mass m,andp→ =df m v→is the momentum.For a collection of objects, whose masses are m k , k = 1, 2, … , nwe haveddt P→tot ≡ ddt ∑ p→ kk=1n = ∑ F→kk=1n ≡ F→tot.In particular, if all the forcesF→k sum to zero1, thetime derivative of the total momentum P→totvan-ishes, meaning that the total momentum is con-stant. This is the content of Newton’s First Law ofMotion,“In the absence of external forces, an object at restremains at rest, and an object in uniform motion re-mains in that state of motion.”Another name for this principle is Conservation ofMomentum.ExampleA car of mass 1000 kg moving at 30 km/hr in thex direction collides with a car of mass 2000 kgmoving at 20 km/hr in the −y direction. What isthe velocity of the resulting mess, ignoring frictionwith the road and assuming the cars stick togetherafter the collision? Solution:The momentum before equals the momentum af-ter (no external forces!)p→in = 1000 ⋅ 302000 ⋅ (−20) = 3000 ⋅ vxvy ≡ p→outSolving we see immediately thatv→out = 10−16.33 km ⁄ hr .End example2. Work and energyNext we may define the work done by a force on amoving body as the force component in the direc-tion of motion, times the distance moved. Forsmall (vector) displacements dx→ we therefore havedW =df F→ ⋅ dx→ =df Fx dx + Fy dy + Fz dz .If we replace F→ by Newton’s Second Law, and notethat for a moving object the displacement in timedt isPhysics of the Human BodyChapter 1: Review of Newtonian Physics 11. …in a vectorial sense.dx→ = v→ dtwe may integrate ∫ dW = ∫ F→ ⋅ dx→ = ∫ mdv→dt ⋅ dx→ ≡ m ∫ dv→ ⋅ v→= ∫ d 12m v→ ⋅ v→ ≡ 12 m v2 ,thereby finding that the work done in acceleratingan object from zero velocity to velocity v→ isW = 12 m v→ ⋅ v→ ≡ 12 m v2 .The latter expression is called the kinetic energy ofa moving object because it is strictly associatedwith the object’s motion.3. PowerFrom the expressiondW = F→ ⋅ dx→for work we note that the energy consumed perunit time in exerting a force upon a moving bodyisP =df dWdt = F→ ⋅ dx→dt ≡ F→ ⋅ v→ .The rate of energy expenditure per unit time iscalled power.Examples of power consumption are found in pro-pulsion problems—we exert a certain force to keepa vehicle or ship moving at constant speed againstvarious resistive forces such as air resistance, vis-cosity, friction, wave generation and the like, thatwould tend to slow it down.ExampleWhat is the total mechanical power required by aman of mass 100 kg, to walk up a 5% grade at3.5 mi/hr, given that walking on the level at thatspeed requires 120 Watts?Solution:A 5% grade rises 5 feet for each 100 feet of forwardprogress. Thus after walking an hour (3600 sec-onds) the man has risen about 282 meters, therebydoing total mechanical workW = 100 kg × 9.8 m ⁄ sec2 × 282 m= 276360 Joules .His power consumption from climbing is therefore∆P = 276360 J ⁄ 3600 sec = 77 Watts ,hence his total is almost 200 Watts.End of Example4. Extended bodiesCenter of massUp until now we have been pretending that massesare mathematical points, with no extension. Anymotion of an extended body—which for now weshall consider rigid—can be decomposed into twoindependent motions: displacement of the centerof mass, and rotation about the center of mass.We begin by defining the center of mass: an ex-tended body can be thought of as comprised of Nindividual pointlike particles, each of mass m k .The center of mass is a point in space (it need notbe located within the body) defined in terms of thepositions of the individual locations of its compo-nents:X→cm =df ∑ m k x→kk=0N∑ m kk=1NThe importance of X→cmis that the force of gravi-tation on an object behaves as though all the masscomprising it were concentrated at the center ofmass. Moreover, if the sum of all the external forcesacting on each of the N constituents does not2 Chapter 1: Review of Newtonian Physicsvanish, the equation of motion describing the dis-placement of an extended object isF→tot = M d2X→cmdt2,whereM = ∑ m kk=1N .RotationBecause displacements and rotations are inde-pendent we may treat rotations as though a givenobject’s center of mass were at rest. Now for thisto continue to be so, the net force acting on thebody must vanish. One way this can happen, how-ever, is for equal and opposite forces to act ondifferent parts of the object, as shown below:Naturally we expect the unbalanced forces shownto cause the object to rotate about its center in theclockwise direction, and indeed that is what willhappen.In order to describe rotations in terms of vectorswe must define a new way to multiply two vectors.Presumably the reader is already familiar with thescalar, or “dot” product of two vectors:a→ ⋅ b→ =df axbx + ayby + azbz ,that we used in defining work and power. This iscalled a scalar product because the result is just anumber; moreover, if the coordinate system is ro-tated the scalar product (computed in terms ofcomponents along the new coordinate axes) is thesame number. We say that the scalar product isinvariant with respect to rotation of coordinates.The proof of this statement is so is so simpleand clever that I cannot resist including it.Consider the scalar product of a vector with it-self: a→ ⋅ a→ = axax + ayay + azazFrom the Pythagorean Theorem we see thatthis is just the square of the length of the vec-tor, which is obviously unchanged by a rota-tion. Since the (squared) length of a vector isinvariant, so must be the difference of the(squared) lengths of two vectors, or any con-stant multiple thereof: that is,14 a→ + b→ ⋅ a→ + b→ − a→ − b→ ⋅ a→ − b→≡ a→ ⋅ b→is invariant under rotations.A final remark about the scalar product: if wedenote the length of a vector bya→ = √ a→ ⋅ a→ ≡ athen the scalar product has the valuea→ ⋅ b→ = a b cosθwhere θ is the angle between the two vectors.The new way to multiply two vectors is called thevector product (or sometimes “cross” product) andis defined as