. Peraire 16.07 Dynamics Ve Lecture D19-2D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture D7 for particle dynamics and extend it to 2D rigid body dynamics. Kinetic Energy for a 2D Rigid Body We start by recalling the kinetic energy expression for a system of particles derived in lecture D17, y G G T v2+ There n is the total number of particles, mi denotes the mass of particle i, and ri is the position vector of particle i with respect to the center of mass, G. Also, m=2ia mi is the total mass of the system, and UG is the velocity of the center of mass. The above expression states that the kinetic energy of a system of particles equals the kinetic energy of a particle of mass m moving with the velocity of the center of mass plus the kinetic energy due to the motion of the particles relative to the center of mass, G. For a 2D rigid body, the velocity of all particles relative to the center of mass is a pure rotation. Thus, we can write r1=w×r remore 42=∑2m(xn)间x)=∑2mn22 here we have used the fact that w and r are perpendicular. The term 2i, mir/2 is ea lized as he moment of inertia, IG, about the center of mass, G. Therefore, for a 2D rigid body, the kinetic energy simply
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D19 - 2D Rigid Body Dynamics: Work and Energy In this lecture, we will revisit the principle of work and energy introduced in lecture D7 for particle dynamics, and extend it to 2D rigid body dynamics. Kinetic Energy for a 2D Rigid Body We start by recalling the kinetic energy expression for a system of particles derived in lecture D17, T = 1 2 mv2 G + Xn i=1 1 2 mir˙ ′ i 2 , where n is the total number of particles, mi denotes the mass of particle i, and r ′ i is the position vector of particle i with respect to the center of mass, G. Also, m = Pn i=1 mi is the total mass of the system, and vG is the velocity of the center of mass. The above expression states that the kinetic energy of a system of particles equals the kinetic energy of a particle of mass m moving with the velocity of the center of mass, plus the kinetic energy due to the motion of the particles relative to the center of mass, G. For a 2D rigid body, the velocity of all particles relative to the center of mass is a pure rotation. Thus, we can write r˙ ′ i = ω × r ′ i . Therefore, we have Xn i=1 1 2 mir˙ ′ i 2 = Xn i=1 1 2 mi(ω × r ′ i ) · (ω × r ′ i ) = Xn i=1 1 2 mir ′ i 2ω 2 , where we have used the fact that ω and r ′ i are perpendicular. The term Pn i=1 mir ′ i 2 is easily recognized as the moment of inertia, IG, about the center of mass, G. Therefore, for a 2D rigid body, the kinetic energy is simply, T = 1 2 mv2 G + 1 2 IGω 2 . (1) 1
When the body is rotating about a fixed point O, we can write Io= IG +mr2 and T==mu2+=(o-mrG)o-2 sInce UG wr The above expression is also applicable in the more general case when there is no fixed point in the motion provided that O is replaced by the instantaneous center of rotation. Thus, in general We shall see that, when the instantaneous center of rotation is known, the use of the above expression does simplify the algebra considerably. Work Recall that the work done by a force F, over an infinitesimal displacement, dr, is dW=F. dr. If Fto denotes the resultant of all forces acting on particle i, then we can write dw:= ftot du Ir:=mi dt. dr:=miU, dv=d( mv2)=d(T), where we have assumed that the velocity is measured relative to an inertial reference frame, and, hence Ftotal= mia;. The above equation states that the work done on particle i by the resultant force Tota is equal to the change in its kinetic energy The total work done on particle i, when moving from position 1 to position 2, is (Wi)1-2 dwi and, summing over all particles, we obtain the principle of work and energy for syst T1+∑(W)1-2=T2 The force acting on each particle will be the sum of the internal forces caused by the other particles, and the erternal forces. We now consider separately the work done by the internal and external forces Internal forces We shall assume, once again, that the internal forces due to interactions between particles act along the lines joining the particles, thereby satisfying Newton's third law. Thus, if fii denotes the force that particle j exerts on particle i, we have that fi is parallel to ri-Ty, and satisfies fi=-fj i Let us now look at two particles, i and j, undergoing an infinitesimal rigid body motion, and consider the f;·mr+f·dr
When the body is rotating about a fixed point O, we can write IO = IG + mr2 G and T = 1 2 mv2 G + 1 2 (IO − mr2 G)ω 2 = 1 2 IOω 2 , since vG = ωrG. The above expression is also applicable in the more general case when there is no fixed point in the motion, provided that O is replaced by the instantaneous center of rotation. Thus, in general, T = 1 2 IC ω 2 . We shall see that, when the instantaneous center of rotation is known, the use of the above expression does simplify the algebra considerably. Work Recall that the work done by a force, F, over an infinitesimal displacement, dr, is dW = F · dr. If F total i denotes the resultant of all forces acting on particle i, then we can write, dWi = F total i · dri = mi dvi dt · dri = mivi · dvi = d( 1 2 miv 2 i ) = d(Ti) , where we have assumed that the velocity is measured relative to an inertial reference frame, and, hence, F total i = miai . The above equation states that the work done on particle i by the resultant force F total i is equal to the change in its kinetic energy. The total work done on particle i, when moving from position 1 to position 2, is (Wi)1−2 = Z 2 1 dWi , and, summing over all particles, we obtain the principle of work and energy for systems of particles, T1 + Xn i=1 (Wi)1−2 = T2 . (2) The force acting on each particle will be the sum of the internal forces caused by the other particles, and the external forces. We now consider separately the work done by the internal and external forces. Internal Forces We shall assume, once again, that the internal forces due to interactions between particles act along the lines joining the particles, thereby satisfying Newton’s third law. Thus, if fij denotes the force that particle j exerts on particle i, we have that fij is parallel to ri − rj , and satisfies fij = −fji. Let us now look at two particles, i and j, undergoing an infinitesimal rigid body motion, and consider the term, fij · dri + fji · drj . (3) 2
If we write dr= dri +d(ri-ri), then, "i)-fiid(Ti-Ti)=-fii d(Tj-Ti) It turns out that fi; d(r-ri) is zero, since d(ri-ri) is perpendicular to ry -ri, and hence it is also perpendicular to fii. The orthogonality between d(Ti-Ti)and -Ti follows from the fact that the distance between any two particles in a rigid body must remain constant(i.e(r) -Ti).(T; -ri)=(;-ri)2 thus differentiating, we have 2d(ri-Ti)(Ti-Ti)=0) d We conclude that, since all the work done by the internal forces can be written as a sum of terms of the form 3, then the contribution of all the internal forces to the term 2ia(wi)1-2 in equation 2, is zero External forces We have established in the previous section that only the external forces to the rigid body are capable of doing any work. Thus, the total work done on the body will b (Wi)1 where Fi is the sum of all the external forces acting on partick Work done by couples If the sum of the external forces acting on the rigid body is zero, it is still possible to have non-zero work. Consider, for instance, a moment M= Fa acting on a rigid body. If the body undergoes a pure translation it is clear that all the points in the body experience the same displacement, and, hence, the total work done by a couple is zero. On the other hand, if the body experiences a rotation de, then the work done by the dw= F-do+F-do= F ado= Mde de F If M is constant, the work is simply W1-2=M(02-0l)
If we write drj = dri + d(rj − ri), then, fij · dri + fji · drj = fij · (dri − dri) − fij · d(rj − ri) = −fij · d(rj − ri). It turns out that fij · d(rj − ri) is zero, since d(rj − ri) is perpendicular to rj − ri , and hence it is also perpendicular to fij . The orthogonality between d(rj −ri) and rj −ri follows from the fact that the distance between any two particles in a rigid body must remain constant (i.e. (rj −ri)·(rj −ri) = (rj −ri) 2 = const; thus differentiating, we have 2d(rj − ri) · (rj − ri) = 0). We conclude that, since all the work done by the internal forces can be written as a sum of terms of the form 3, then the contribution of all the internal forces to the term Pn i=1(Wi)1−2 in equation 2, is zero. External Forces We have established in the previous section that only the external forces to the rigid body are capable of doing any work. Thus, the total work done on the body will be Xn i=1 (Wi)1−2 = Xn i=1 Z (ri)2 (ri)1 Fi · dr , where Fi is the sum of all the external forces acting on particle i. Work done by couples If the sum of the external forces acting on the rigid body is zero, it is still possible to have non-zero work. Consider, for instance, a moment M = F a acting on a rigid body. If the body undergoes a pure translation, it is clear that all the points in the body experience the same displacement, and, hence, the total work done by a couple is zero. On the other hand, if the body experiences a rotation dθ, then the work done by the couple is dW = F a 2 dθ + F a 2 dθ = F adθ = M dθ . If M is constant, the work is simply W1−2 = M(θ 2 − θ 1 ) 3
Conservative forces When the forces can be derived from a potential energy function, v, we say e forces are con such cases, we have that F=-Vv, and the work and energy relation in equation 2 takes a particularly simple form. Recall that a necessary, but not sufficient, condition for a force to be conservative is that it must be a function of position only, i. e. F(r)and V(r). Common examples of conservative forces are gravity (a constant force independent of the height), gravitational attraction between two bodies(a force inversely proportional to the squared distance between the bodies), and the force of a perfectly elastic spn: sely The work done by a conservative force between position ri and r2 is Thus, if we call Wis the work done by all the external forces which are non conservative, we can write the Ti+Vi+Wi=T2 Of course, if all the forces that do work are conservative, we obtain conservation of total energy, which can Gravity Potential for a Rigid Body In this case, the potential Vi associated with particle i is simply Vi=migzi, where zi is the height of particle i above some reference height. The force acting on particle i will then be Fi=-VVi. The work done on the whole body will be f∫ V)1-(V)2)=∑mg(2)-(2 There the gravity potential for the rigid body is simply, mi92i=920 where zg is the z coordinate of the center of mass. Example Cylinder on a Ramp We consider a homogeneous cylinder released from rest at the top of a ramp of angle o, and use conservation of energy to derive an expression for the velocity of the cylinder
Conservative Forces When the forces can be derived from a potential energy function, V , we say the forces are conservative. In such cases, we have that F = −∇V , and the work and energy relation in equation 2 takes a particularly simple form. Recall that a necessary, but not sufficient, condition for a force to be conservative is that it must be a function of position only, i.e. F(r) and V (r). Common examples of conservative forces are gravity (a constant force independent of the height), gravitational attraction between two bodies (a force inversely proportional to the squared distance between the bodies), and the force of a perfectly elastic spring. The work done by a conservative force between position r1 and r2 is W1−2 = Z r2 r1 F · dr = [−V ] r2 r1 = V (r1) − V (r2) = V1 − V2 . Thus, if we call WNC 1−2 the work done by all the external forces which are non conservative, we can write the general expression, T1 + V1 + WNC 1−2 = T2 + V2 . Of course, if all the forces that do work are conservative, we obtain conservation of total energy, which can be expressed as, T + V = constant . Gravity Potential for a Rigid Body In this case, the potential Vi associated with particle i is simply Vi = migzi , where zi is the height of particle i above some reference height. The force acting on particle i will then be Fi = −∇Vi . The work done on the whole body will be Xn i=1 Z r 2 i r 1 i fi · dri = Xn i=1 ((Vi)1 − (Vi)2) = Xn i=1 mig((zi)1 − (zi)2 = V1 − V2 , where the gravity potential for the rigid body is simply, V = Xn i=1 migzi = mgzG , where zG is the z coordinate of the center of mass. Example Cylinder on a Ramp We consider a homogeneous cylinder released from rest at the top of a ramp of angle φ, and use conservation of energy to derive an expression for the velocity of the cylinder. 4
S Conservation of energy implies that T +V= Tinitial+Initial. Initially, the kinetic energy is zero, Tinitial Thus, for a later time, the kinetic energy is given by T=Initial -V= mgs sin o where s is the distance traveled down the ramp. The kinetic energy is simply T=i IG + mR2 is the moment of inertia about the instantaneous center of rotation C, and w is the angular 1+(Ic/mR2) since W=U/R Equilibrium and Stability If all the forces acting on the body are conservative, then the potential energy can be used very effectively to determine the equilibrium positions of a system and the nature of the stability at these positions. Let us assume that all the forces acting on the system can be derived from a potential energy function, v. It is clear that if F=-VV=0 for some position, this will be a point of equilibrium in the sense that if the body is at rest(kinetic energy zero), then there will be no forces(and hence, no acceleration) to change the equilibrium, since the resultant force F is zero. Once equilibrium has been established the stability of he equilibrium point can be determine by examining the shape of the potential function. If the potential function has a minimum at the equilibrium point, then the equilibrium will be stable. This means that if the potential energy is at a minimum, there is no potential energy left that can be traded for kinetic energy. Analogously, if the potential energy is at a marimum, then the equilibrium point is unstable. Example Equilibrium and Stability A cylinder of radius R, for which the center of gravity, G, is at a distance d from the geometric center, C lies on a rough plane inclined at an angle o
Conservation of energy implies that T +V = Tinitial+Vinitial. Initially, the kinetic energy is zero, Tinitial = 0. Thus, for a later time, the kinetic energy is given by T = Vinitial − V = mgs sin φ , where s is the distance traveled down the ramp. The kinetic energy is simply T = 1 2 IC ω 2 , where IC = IG + mR2 is the moment of inertia about the instantaneous center of rotation C, and ω is the angular velocity. Thus, IC ω 2 = 2mgs sin φ , or, v 2 = 2gs sin φ 1 + (IG/mR2) , since ω = v/R. Equilibrium and Stability If all the forces acting on the body are conservative, then the potential energy can be used very effectively to determine the equilibrium positions of a system and the nature of the stability at these positions. Let us assume that all the forces acting on the system can be derived from a potential energy function, V . It is clear that if F = −∇V = 0 for some position, this will be a point of equilibrium in the sense that if the body is at rest (kinetic energy zero), then there will be no forces (and hence, no acceleration) to change the equilibrium, since the resultant force F is zero. Once equilibrium has been established, the stability of the equilibrium point can be determine by examining the shape of the potential function. If the potential function has a minimum at the equilibrium point, then the equilibrium will be stable. This means that if the potential energy is at a minimum, there is no potential energy left that can be traded for kinetic energy. Analogously, if the potential energy is at a maximum, then the equilibrium point is unstable. Example Equilibrium and Stability A cylinder of radius R, for which the center of gravity, G, is at a distance d from the geometric center, C, lies on a rough plane inclined at an angle φ. 5
Since gravity is the only external force acting on the cylinder that is capable of doing any work, we can ne the equilibrium and stability of the system by considering the potential energy function. We have ZC0-RO sin o, where aco is the value of zc when 0=0. Thus, since d=CG, we have V=mgaG= mg(2c +dsin 0)=mg(aco- RO sin o +dsin 0) The equilibrium points are given by VV=0, but, in this case, since the position of the system is uniquely determined by a single coordinate, e.g. 0, we can write dv/de mg(-Rsin+d cos 8)=0, or, cos 0=(Rsin o)/d. If d Rsin o, then ge cos[(Rsin o)/d is an equilibrium point. We note that if geq. is an equilibrium point, then -Beg. is also an equilibrium point (i.e. cos 0= cos(-0) In order to study the stability of the equilibrium points, we need to determine whether the potential energy is a maximum or a minimum at these points. Since dv/d82= - sin 0, we have that when geq. 0 and the potential energy is a minimum at that point. Consequently, for geq.0, the equilibrium point is unstable ADDITIONAL READING J. L. Meriam and L G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition
Since gravity is the only external force acting on the cylinder that is capable of doing any work, we can examine the equilibrium and stability of the system by considering the potential energy function. We have zC = zC0 − Rθ sin φ, where zC0 is the value of zC when θ = 0. Thus, since d = |CG|, we have, V = mgzG = mg(zC + d sin θ) = mg(zC0 − Rθ sin φ + d sin θ) . The equilibrium points are given by ∇V = 0, but, in this case, since the position of the system is uniquely determined by a single coordinate, e.g. θ, we can write ∇V = dV dθ ∇θ , which implies that, for equilibrium, dV /dθ = mg(−R sin φ + d cos θ) = 0, or, cos θ = (R sin φ)/d. If d 0 and the potential energy is a minimum at that point. Consequently, for θ eq. 0, the equilibrium point is unstable. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 6/6, 6/7 6
