16.07 Dynamics Fall 2004 Ⅴ ersIon1.1 Lecture D8-Conservative Forces and Potential Energy We have seen that the work done by a force F on a particle is given by dw= F. dr. If the work done by F, when the particle moves from any position r to any position T2, can be expressed 2=F=()-V()=1-1, then we say that the force is conservative. In the above expression, the scalar function V(r) is called the potential. It is clear that the potential satisfies dV=-F. dr(the minus sign is included for convenience) There are two main consequences that follow from the existence of a potential: i)the work done by a conservative force between points r1 and r2 is independent of the path. This follows from(1) since W12 only depends on the initial and final potentials Vi and V2(and not on how we go from r to r2),and ii)the work done by potential forces is recoverable. Consider the work done in going from point ri to W12. If we go, now, from point r2 to ri, we have that W2l ==w12 since the total work W12+ one nsion any force which is only a function of position is conservative. That is, if we have a force, F(a), which is only a function of position, then F(ar)dr is always a perfect differential. This means that we can define a potential function as (x)=-/F()dr here ro is arbitrary In two and three dimensions, we would, in principle, expect that any force which depends only on position F(r), to be conservative. However, it turns out that, in general, this is not sufficient. In multiple dimensions, he condition for a force field to be conservative is that it can be expressed as the gradient of a potential function. That is Note The gradient operator, V The gradient operator, V(called"del"), in cartesian coordinates is defined as a( VO=02+0uj+ azk
J. Peraire 16.07 Dynamics Fall 2004 Version 1.1 Lecture D8 - Conservative Forces and Potential Energy We have seen that the work done by a force F on a particle is given by dW = F · dr. If the work done by F, when the particle moves from any position r1 to any position r2, can be expressed as, W12 = Z r2 r1 F · dr = −(V (r2) − V (r1)) = V1 − V2 , (1) then we say that the force is conservative. In the above expression, the scalar function V (r) is called the potential. It is clear that the potential satisfies dV = −F · dr (the minus sign is included for convenience). There are two main consequences that follow from the existence of a potential: i) the work done by a conservative force between points r1 and r2 is independent of the path. This follows from (1) since W12 only depends on the initial and final potentials V1 and V2 (and not on how we go from r1 to r2), and ii) the work done by potential forces is recoverable. Consider the work done in going from point r1 to point r2, W12. If we go, now, from point r2 to r1, we have that W21 = −W12 since the total work W12 + W21 = (V1 − V2) + (V2 − V1) = 0. In one dimension any force which is only a function of position is conservative. That is, if we have a force, F(x), which is only a function of position, then F(x) dx is always a perfect differential. This means that we can define a potential function as V (x) = − Z x x0 F(x) dx , where x0 is arbitrary. In two and three dimensions, we would, in principle, expect that any force which depends only on position, F(r), to be conservative. However, it turns out that, in general, this is not sufficient. In multiple dimensions, the condition for a force field to be conservative is that it can be expressed as the gradient of a potential function. That is, F = −∇V . Note The gradient operator, ∇ The gradient operator, ∇ (called “del”), in cartesian coordinates is defined as ∇( ) ≡ ∂( ) ∂x i + ∂( ) ∂y j + ∂( ) ∂z k . 1
When operating on a scalar function V(r, y, a), the result VV is a vector, called the gradient of V. The components of VV are the derivatives of v along each of the coordinate directions VI or If we consider a particle moving due to conservative forces with potential energy V(a, y, 2), as the particle moves from point r=ri+y3 2k to point r+dr =(a +dr)i+(y+dy)3+(2+dz)k, the potential energy changes by dV v(a + dr, y+dy, 2 + dz)-v(r, 3, 2). For small increments dr, dy, dz, and dv, can be expressed, using Taylor series expansions, as avr+ o dy+az dz= vv. dr dsk This equation expresses the fundamental property of the gradient. The gradient allows us to find the change in a function induced by a change in its variables If we write V(, y, 2)=C, for some constant C, this is the implicit equation of a surface, which is called a constant energy surface. This surface is made up by all the points in the z, y, a space for which the function v(, y, z)is equal to C. It is clear that if a particle moves on a constant energy surface, dV=0, since V is constant on that surface. Therefore, when a particle moves on a constant energy surface, dr will be tangent to that surface. and since 0=dv=vV·dr, re have that vv is perpendicular to any tangent to the surface. This situation is illustrated in the picture elow for the two dimensional case. Here, the constant energy surfaces are contour curves, and we can see that the gradient vector is always normal to the contour curves v(e, y, z)=C1 VO Gradient operator in cylindrical coordinates The gradient operator can be expressed in cylindrical coordinates by writing a= r cos 0, y= rsin 8, and
When operating on a scalar function V (x, y, z), the result ∇V is a vector, called the gradient of V . The components of ∇V are the derivatives of V along each of the coordinate directions, ∇V ≡ ∂V ∂x i + ∂V ∂y j + ∂V ∂z k . If we consider a particle moving due to conservative forces with potential energy V (x, y, z), as the particle moves from point r = xi + yj + zk to point r + dr = (x + dx)i + (y + dy)j + (z + dz)k, the potential energy changes by dV = V (x + dx, y + dy, z + dz) − V (x, y, z). For small increments dx, dy, dz, and dV , can be expressed, using Taylor series expansions, as dV = ∂V ∂x dx + ∂V ∂y dy + ∂V ∂z dz = ∇V · dr , where dr = dxi + dyj + dzk. This equation expresses the fundamental property of the gradient. The gradient allows us to find the change in a function induced by a change in its variables. If we write V (x, y, z) = C, for some constant C, this is the implicit equation of a surface, which is called a constant energy surface. This surface is made up by all the points in the x, y, z space for which the function V (x, y, z) is equal to C. It is clear that if a particle moves on a constant energy surface, dV = 0, since V is constant on that surface. Therefore, when a particle moves on a constant energy surface, dr will be tangent to that surface, and since 0 = dV = ∇V · dr , we have that ∇V is perpendicular to any tangent to the surface. This situation is illustrated in the picture below for the two dimensional case. Here, the constant energy surfaces are contour curves, and we can see that the gradient vector is always normal to the contour curves. Note Gradient operator in cylindrical coordinates The gradient operator can be expressed in cylindrical coordinates by writing x = r cos θ, y = r sin θ, and 2
r=V22+y2, 0=tan1(y/z). Thus, applying the chain rule for differentiation, we have 0() dr a( ar a0 0()sin6O() 0() ar( arao 0(),cos60() If we note that i= cos Ber- sin Beg and j= sin Ber +cos Bee, we have that 00。10。0 r 00 An expression for spherical coordinates can be derived in a similar manner Conservation of Energy When all the forces doing work are conservative, the work is given by(1), and the principle of work and energy derived in the last lecture, T1+W12=T2 reduces to 11+V=T2+V2 or more generally, since the points ri and r2 are arbitrary, E=T+V Whenever applicable, this equation states that the total energy stays constant, and that during the motion In the general case, however, we will have a combination of conservative, F, and non-conservative, F forces. In this case, the work done by the conservative forces will be calculated using the corresponding potential function, i.e., W9= Vi-V2, and the work done by the non-conservative forces will be path dependent and will need to be be calculated using the work integral. Thus, in the general case, we will have T1+V1+ FNC,dr Examples of Conservative Forces Gravity near the earth's surface On a"fat earth", the specific gravity g points down(along the-z axis), so F=-mgk. Call V =0 on the surface 2=0. and then v(a) mg2
r = p x 2 + y 2, θ = tan−1 (y/x). Thus, applying the chain rule for differentiation, we have ∂( ) ∂x = ∂r ∂x ∂( ) ∂r + ∂r ∂x ∂( ) ∂θ = cos θ ∂( ) ∂r − sin θ r ∂( ) ∂r ∂( ) ∂y = ∂r ∂y ∂( ) ∂r + ∂r ∂y ∂( ) ∂θ = sin θ ∂( ) ∂r + cos θ r ∂( ) ∂r . If we note that i = cos θer − sin θeθ and j = sin θer + cos θeθ, we have that ∇( ) ≡ ∂( ) ∂r er + 1 r ∂( ) ∂θ eθ + ∂( ) ∂z . An expression for spherical coordinates can be derived in a similar manner. Conservation of Energy When all the forces doing work are conservative, the work is given by (1), and the principle of work and energy derived in the last lecture, T1 + W12 = T2 , reduces to, T1 + V1 = T2 + V2 or more generally, since the points r1 and r2 are arbitrary, E = T + V = constant . (2) Whenever applicable, this equation states that the total energy stays constant, and that during the motion only exchanges between kinetic and potential energy occur. In the general case, however, we will have a combination of conservative, F C , and non-conservative, F NC , forces. In this case, the work done by the conservative forces will be calculated using the corresponding potential function, i.e., WC 12 = V1 − V2, and the work done by the non-conservative forces will be path dependent and will need to be be calculated using the work integral. Thus, in the general case, we will have, T1 + V1 + Z r2 r1 F NC · dr = T2 + V2 . Examples of Conservative Forces Gravity near the earth’s surface On a “flat earth”, the specific gravity g points down (along the -z axis), so F = −mgk. Call V = 0 on the surface z = 0, and then V (z) = − Z z 0 (−mg) dz, V (z) = mgz . 3
For the motion of a projectile, the total energy is then 1 mu+ mgz= constant Since v and v, remain constant, we also have -muz +mgz= constant gravity In a central gravity field and so, taking I(r→∞)=0, Mn Spring Fo orce The force supported by a spring is F==kr. The elastic potential energy of the spring is the work done on it to deform it an amount z. Thus, we have -kr dxr If the deformation, either tensile or compressive, increases from 1 to z2 during the motion, then the change in potential energy of the spring is the difference between its final and initial values, or △V=k(x2-x2) Exercise Non-conservative forces Consider a block sliding on a horizontal surface. Is the friction force conservative? Explain why? Give examples of non-conservative forces Note Path independent forces for which there is no potential(Optional) The condition F=-VV is sufficient, bu ut not necessa ary for the work to be path independent. An important example is the magnetostatic force F=qU x B, where the work F.dr= F. udt is zero(the force is alway perpendicular to u). Thus, for a general velocity-dependent electro-magnetic force, F=q(E+uxB),where only B is irrotational, the work is still path independent, and the total energy(mu2/2+qo)is conserve
For the motion of a projectile, the total energy is then E = 1 2 mv2 + mgz = constant . Since vx and vy remain constant, we also have 1 2 mv2 z + mgz = constant. Gravity In a central gravity field F = −G Mm r 2 er = −∇(−G Mm r ) , and so, taking V (r → ∞) = 0, V = −G Mm r . Spring Force The force supported by a spring is F = −kx. The elastic potential energy of the spring is the work done on it to deform it an amount x. Thus, we have V = − Z x 0 −kx dx = 1 2 kx2 . If the deformation, either tensile or compressive, increases from x1 to x2 during the motion, then the change in potential energy of the spring is the difference between its final and initial values, or, ∆V = 1 2 k(x 2 2 − x 2 1 ) . Exercise Non-conservative Forces Consider a block sliding on a horizontal surface. Is the friction force conservative? Explain why? Give examples of non-conservative forces. Note Path independent forces for which there is no potential (Optional) The condition F = −∇V is sufficient, but not necessary for the work to be path independent. An important example is the magnetostatic force F = qv × B, where the work F · dr = F · vdt is zero (the force is always perpendicular to v). Thus, for a general velocity-dependent electro-magnetic force, F = q(E +v×B), where only B is irrotational, the work is still path independent, and the total energy (mv2/2 + qφ) is conserved. 4
Re ferenc 1 M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96 ADDITIONAL READING J. L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 7
References [1] M. Martinez-Sanchez, Unified Engineering Notes, Course 95-96. ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 3/7 5
