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Work and Energy 1m 4 kg LAB = mgh 1 kg 1 kg LAB = 4( m 4 gh) = mgh LAB = mgl sin α = mgd hd = mgh Work of a force: L = F� · �s = F s cos θ = F � s = [N m] = [J] = [Joule] W = work done by a constant force F� to move a body by a displacement �s. θ = angle between the vectors F� and �s (F � is the component of F� parallel to �s). � (�r) = Fx (�r)î + Fy (�r)ĵ + Notice: For a non constance force F Fz (�r)k̂ one needs to calculate the in work, i.e. the work for an infinitesimal displacement d�r� = dxî + dy ĵ + dz k̂ and sum over B � the body path A → B: L = A F (�r) · d�r F� � The contact forces (e.g. normal force) do not contribute to the work! (by definition they are perpendicular to the displacement) For conservative forces the work does not depends on the path between the initial configuration and the final configuration of the body. (“the work does not depend on how the work is done”) Work and Energy For conservative forces the work does not depends on the path between the initial configuration and the final configuration of the body. (“the work does not depend on how the work is done”) Energy: The Energy is the capacity of a body to do a work, it is expressed in [Joule] (the variation of energy is the work done by the force acting on the body). Potential Energy LAB = −(UB − UA ) = −∆U the potential energy of a conservative force is the energy associated to the position or to the state of a body subject to that force LAB = mgl sin α = mgd hd = mgh m m m/4 h l h = l sin α α LAB = mgh LAB = 4( m 4 gh) = mgh Work and Energy Gravitational potential energy (on Earth surface) Ug = mgh Ug = gravitational potential energy on the Earth surface of a body on mass m at height y = h relative to a reference position y = 0 (h � rT where rT = Earth radius) LAB = mgl sin α = mgd hd = mgh m m m/4 l h = l sin α α LAB = mgh LAB = 4( m 4 gh) = mgh Work and Energy Kinetic energy 1 2 K= mv 2 K = energy of a body m moving at velocity v (= work produce to put the body at rest). The total work LAB done on the body that moves from position A to position B is equal to the variation of its kinetic energy: LAB = KB − KA F� � = Elastic potential energy �vA � a� m �vB 1 2 kx 2 Uk = elastic potential energy of a body attached to a spring displaced of x from the equilibrium condition; k = elastic constant of the spring. Uk = Interpretation of the potential energy graphic A body subject to conservative forces moves toward the minimum of the corresponding potential energy U (ti ) + K(ti ) = U (t) + K(t) = U (tf ) + K(tf ) , ∀t . . x(t) Ug (ti ) Uk (ti ) K(t) K(t) UK (t) Ug (t) Equilibrium position . h(t) Principle of conservation of Mechanical Energy In a ISOLATED system the TOTAL energy is conserved (it is constant in time), but it may transforms from a type of energy to the other. For a system on which on conservative forces act ∆K + ∆Ug + ∆Uk + ∆Ualtro = 0 where: ∆K = variation of kinetic energy of the system; ∆Ug = variation of gravitational energy; ∆Uk variation of elastic energy; ∆Uother variation of other types of conservative potential energy The energy can not created nor distroied For dissipative forces (e.g.: friction) the corresponding work done depends on the particular path and it is not possible to define a potential. In this case part of the energy of the system is converted in “internal energy” (e.g.: Heat) of the system and it can not be directly converted to work. Gravitational potential energy (general case) m1 · m2 UG = −G r Gravitational potential energy of a point-like particle of mass m1 at distance r from a second point-like particle of mass m2 (notice: by convention r → ∞ ⇒ UG → 0). Molecular bounding energy Power Variation of work over time: Pm W = t joule [ ≡ W AT T = W ] second Pm = mean power of a force is equal to the work done W divided by the time t taken for doing it. Another common unit for the Power is the HOURSEPOWER (hp): 1hp = 746W Notice: the Kilowatt-hour (KW h) is 1kW × h = 3.6 × 106 J = 3.6M J. It denotes an energy (work). Metabolism promptness of energy use in leaving beings Moment of a force w.r.t. a point (a.k.a torque) �τ = �r × F� → |�τ | = r · F · sin α = F · d = F ⊥ · r [newton · metro → N · m] [joule] � τ F� �r α d = r sin α τ = moment w.r.t. to the point O (fulcrum) of the force F� applied to the point P . d = lever-arm of the force: distance between the rect of application of F� from O. �r = � . α = angle between �r and F� OP Lθ = F1⊥ d1 θ = F2⊥ d2 θ → F1⊥ d1 = F2⊥ d2 = cost �τ1 = �τ2 = �τ = cost �τ2 �τ1 Impulse of force J� = F� · ∆t , [N · s] J� (mean) impuls of the (constant) force F� applied to a body for a �time interval ∆t. Total impulse for non-constant forces t J� = tif F� (t) · dt m �q = m · �v , [kg · ] Momentum s �q = momentum of a body m and velocity �v . J� = ∆�q the impulse J� of a force is is the variation of momentum ∆�q of the body. q The second Newton’s law takes the forme F� = d� dt . The momentum of an isolated system (F� = 0) is conserved �q = cost. Elastic (linear) collision between a particle and a target The elastic collisions between particles are solved by imposing separately the conservation of the energy and of the momentum between the conditions of the system before and after the collision. m1 v1 = m1 v1� + m2 v2� , v1� = 1 1 1 m1 v12 = m1 v1�2 + m2 v2�2 2 2 2 (m1 − m2 )v1 , m1 + m2 v2� = before after 2m1 v1 m1 + m2 Completely inelastic (linear) collision (lineare) with target In the inelastic collision the particle and target have the same final velocity. Part of the energy is dissipated (e.g.: heat). The collision is solved by determining a single unknown variable by means of the conservation of the total momentum m1 v1 = m1 vf� + m2 vf� → vf� = m1 v 1 m1 + m2 Mechanics of systems of particles Centre of mass �rcm = �N i �N i . �rcm mi�ri Mtot cm �N mi xi m1 x1 + m2 x2 + · · · + mN xN xcm = �i N = Mtot i mi �N mi yi m1 y1 + m2 y2 + · · · + mN yN ycm = �i N = Mtot i mi �N mi zi m1 z1 + m2 z2 + · · · + mN zN zcm = �i N = Mtot i mi The coordinates of the centre of mass {xcm , ycm , zcm } are given by the mean value of the positions {xi , yi , zi } of the particles �N i = 1, 2, ·, N weighted by their masses mi . Mtot = i mi : total mass of the system of particles. the gravitational force produces a moment with respect to the centre of mass (a.k.a centre of gravity or equilibrium point). Centre of Mass M · �acm = F�ext M = total (constant) mass of the system; �acm = acceleration of the centre of mass; F�ext = resulting of the external forces of the system. � = M · �vcm Q � = �q1 + �q2 + · · · + �qN = total momentum of the system; Q �vcm = velocity of the centre of mass. Condition of equilibrium A rigid body at rest is in equilibrium if the sum of the external forces acting of the body is zero and the sum of the moment of the forces is zero as well FT ot = 0 → v = 0 τT ot = 0 → ω = 0 Angular momentum |�l| = r sin θ · q , [Kg · m2 /s] �l �q �r �l the angular momentum with respect to a point O of a particle P of momentum q = m�v and angle θ w.r.t. �r = P�O; Moment of inertia of a system of particles w.r.t. an axis I= N � 2 Is the analogous for the rotational motion mi ri of the inertial mass for the linear motion (but it depends on the geometry) i=1 Angular momentum: l = I · w, ( Q = M · v) Moment, a.k.a torque: τ = I · α, ( F = M · a) Work (rotational): W = τ · ∆θ, ( W = F · ∆x) Kinetic energy (rotational): K = 12 Iω 2 , ( K = 12 M v 2 ) θ → angle expressed in RAD; ω = dθ dt → angular velocity; a = dω dt → angular acceleration. θ Moment of inertia of a system of particles w.r.t. an axis I= N � i=1 mi ri2 Is the analogous for the rotational motion of the inertial mass for the linear motion (but it depends on the geometry) Angular momentum: l = I · w, ( Q = M · v) Moment, a.k.a torque: τ = I · α, ( F = M · a) Work (rotational): W = τ · ∆θ, ( W = F · ∆x) Kinetic energy (rotational): K = 12 Iω 2 , ( K = 12 M v 2 ) θ → angle expressed in RAD; ω = dθ dt → angular velocity; a = dω dt → angular acceleration. Mechanical properties of solid bodies In solid bodies the elastic laws (e.g. Hooke’s law) are valid within a finite range of forces (elasticity limit). Above this range the deformation become permanent An ideal elastic body is described by the Hooke’s law F = kx where F is the force acting on the body (F = −kx is the reaction of the body), k is the elastic constant and x is the deformation. Longitudinal and transversal elasticity 1 F σ F Li , ∆R = − Ri E A E A ∆L variation of length resulting from a traction (or compression); ∆L variation of transversal direction resulting from a traction (or compression); F force of longitudinal traction (F > 0) or compression (F < 0); E = [N/m2 ] Young’s parameter (it depends on the material); σ Poisson’s parameter (it depends on the material); Li initial length of the bar; Ri initial transversal dimension of the bar. Ri ∆L = Li F ∆R F Shear force 1F tan α = γA α angle of deformation; γ[N/m2 ] rigidity parameter (it depends on the material); F tangent force acting on the material; A area of the surface; F A shear force ∆L α