Dynamics AE-001 Sharif University-Aerospace Dep. Fall 2004

Rectangular Components

Chapter 2 STATICS OF PARTICLES Forces are vector quantities; they add according to the parallelogram law. The magnitude and direction of the resultant R of two forces P and Q can be determined either graphically or by trigonometry. R P A Q

Any given force acting on a particle can be resolved into two or more components, i.e.., it can be replaced by two or more forces which have the same effect on the particle. A force F can be resolved into two components P and Q by drawing a parallelogram which has F for its diagonal; the components P and Q are then represented by the two adjacent sides of the parallelogram and can be determined either graphically or by trigonometry. Q F A P

F = Fx i + Fy j Fx = F cos q Fy = F sin q Fy tan q = Fx F = Fx + Fy 2 A force F is said to have been resolved into two rectangular components if its components are directed along the coordinate axes. Introducing the unit vectors i and j along the x and y axes, F = Fx i + Fy j y Fx = F cos q Fy = F sin q tan q = Fy Fx Fy = Fy j F j F = Fx + Fy 2 q x i Fx = Fx i

Rx = S Rx Ry = S Ry Ry tan q = R = Rx + Ry 2 Rx When three or more coplanar forces act on a particle, the rectangular components of their resultant R can be obtained by adding algebraically the corresponding components of the given forces. Rx = S Rx Ry = S Ry The magnitude and direction of R can be determined from tan q = Ry Rx R = Rx + Ry 2

Fx = F cos qx Fy = F cos qy Fz = F cos qz Fy Fy qy F F qx Fx Fx Fz Fz B B Fy Fy qy A F A F D D O O qx Fx Fx x x Fz Fz E E C C z z y A force F in three-dimensional space can be resolved into components B Fy F Fx = F cos qx Fy = F cos qy A D O Fz = F cos qz Fx x qz E Fz C z

F = F (cosqx i + cosqy j + cosqz k ) l (Magnitude = 1) The cosines of qx , qy , and qz are known as the direction cosines of the force F. Using the unit vectors i , j, and k, we write Fy j F = F l cos qy j Fx i cos qz k x Fz k F = Fx i + Fy j + Fz k cos qx i z or F = F (cosqx i + cosqy j + cosqz k )

l = cosqx i + cosqy j cos2qx + cos2qy F = Fx + Fy + Fz 2 cosqx = Fx F l (Magnitude = 1) cos qy j l = cosqx i + cosqy j + cosqz k Fy j cos qz k F = F l Since the magnitude of l is unity, we have Fx i x cos2qx + cos2qy + cos2qz = 1 Fz k z cos qx i In addition, F = Fx + Fy + Fz 2 cosqx = Fx F cosqy = Fy F cosqz = Fz F

MN = dx i + dy j + dz k MN 1 l = = ( dx i + dy j + dz k ) d y A force vector F in three-dimensions is defined by its magnitude F and two points M and N along its line of action. The vector MN joining points M and N is N (x2, y2, z2) F dy = y2 - y1 l dz = z2 - z1 < 0 dx = x2 - x1 M (x1, y1, z1) x z MN = dx i + dy j + dz k The unit vector l along the line of action of the force is l = = ( dx i + dy j + dz k ) MN 1 d

F = F l = ( dx i + dy j + dz k ) F d Fdx d Fdy d Fdz d Fx = Fy = Fz = d = dx + dy + dz 2 2 2 N (x2, y2, z2) dy = y2 - y1 A force F is defined as the product of F and l. Therefore, dz = z2 - z1 < 0 dx = x2 - x1 M (x1, y1, z1) x z F = F l = ( dx i + dy j + dz k ) F d From this it follows that Fdx d Fdy d Fdz d Fx = Fy = Fz =

When two or more forces act on a particle in three-dimensions, the rectangular components of their resultant R is obtained by adding the corresponding components of the given forces. Rx = S Fx Ry = S Fy Rz = S Fz The particle is in equilibrium when the resultant of all forces acting on it is zero.

Vector Algebra

V = P x Q V = PQ sin q Q x P = - (P x Q) V = P x Q The vector product of two vectors is defined as Q V = P x Q q P The vector product of P and Q forms a vector which is perpendicular to both P and Q, of magnitude V = PQ sin q This vector is directed in such a way that a person located at the tip of V observes as counterclockwise the rotation through q which brings vector P in line with vector Q. The three vectors P, Q, and V - taken in that order - form a right-hand triad. It follows that Q x P = - (P x Q)

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k It follows from the definition of the vector product of two vectors that the vector products of unit vectors i, j, and k are k i i x i = j x j = k x k = 0 i x j = k , j x k = i , k x i = j , i x k = - j , j x i = - k , k x j = - i The rectangular components of the vector product V of two vectors P and Q are determined as follows: Given P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k The determinant containing each component of P and Q is expanded to define the vector V, as well as its scalar components

P = Px i + Py j + Pz k Q = Qx i + Qy j + Qz k V = P x Q = = Vx i + Vy j + Vz k where Vx = Py Qz - Pz Qy Vy = Pz Qx - Px Qz Vz = Px Qy - Py Qx

MO = r x F MO = rF sin q = Fd Mo The moment of force F about point O is defined as the vector product MO = r x F F where r is the position vector drawn from point O to the point of application of the force F. The angle between the lines of action of r and F is q. O r q d A The magnitude of the moment of F about O can be expressed as MO = rF sin q = Fd where d is the perpendicular distance from O to the line of action of F.

Mx = y Fz - z Fy My = zFx - x Fz Mz = x Fy - y Fx Fy j A (x , y, z ) y j The rectangular components of the moment Mo of a force F are determined by expanding the determinant of r x F. Fx i r Fz k O x i x z k z i x Fx j y Fy k z Fz Mo = r x F = = Mx i + My j + Mzk where Mx = y Fz - z Fy My = zFx - x Fz Mz = x Fy - y Fx

rA/B = xA/B i + yA/B j + zA/B k Fy j A (x A, yA, z A) B (x B, yB, z B) In the more general case of the moment about an arbitrary point B of a force F applied at A, we have Fx i r Fz k O x z i xA/B Fx j yA/B Fy k zA/B Fz MB = rA/B x F = rA/B = xA/B i + yA/B j + zA/B k where xA/B = xA- xB yA/B = yA- yB zA/B = zA- zB and

MB = (xA- xB )Fy + (yA- yB ) Fx In the case of problems involving only two dimensions, the force F can be assumed to lie in the xy plane. Its moment about point B is perpendicular to that plane. It can be completely defined by the scalar Fy j A rA/B Fx i (yA - yB ) j B (xA - xB ) i O x z MB = MB k MB = (xA- xB )Fy + (yA- yB ) Fx The right-hand rule is useful for defining the direction of the moment as either into or out of the plane (positive or negative k direction).

P Q = PQ cos q P Q = Px Qx + Py Qy + Pz Qz The scalar product of two vectors P and Q is denoted as P Q ,and is defined as Q q P Q = PQ cos q where q is the angle between the two vectors P The scalar product of P and Q is expressed in terms of the rectangular components of the two vectors as P Q = Px Qx + Py Qy + Pz Qz

POL = Px cos qx + Py cos qy + Pz cos qz The projection of a vector P on an axis OL can be obtained by forming the scalar product of P and the unit vector l along OL. qy A l P qx x O qz z POL = P l Using rectangular components, POL = Px cos qx + Py cos qy + Pz cos qz

Sx Px Qx Sy Py Qy Sz Pz Qz S (P x Q ) = The mixed triple product of three vectors S, P, and Q is Sx Px Qx Sy Py Qy Sz Pz Qz S (P x Q ) = The elements of the determinant are the rectangular components of the three vectors.

l x x Fx l y y Fy l z z Fz MOL = l MO = l (r x F) = y The moment of a force F about an axis OL is the projection OC on OL of the moment MO of the force F. This can be written as a mixed triple product. MO F C l A (x, y, z) r O x z l x x Fx l y y Fy l z z Fz MOL = l MO = l (r x F) = lx, ly , lz = direction cosines of axis OL x, y , z = components of r Fx, Fy , Fz = components of F

Kinematics of Particles Chapter 11

dx dt v = dv dt d 2x dt 2 a = a = dv dx a = v O P x x The velocity v of the particle is equal to the time derivative of the position coordinate x, dx dt v = and the acceleration a is obtained by differentiating v with respect to t, dv dt d 2x dt 2 a = a = or we can also express a as dv dx a = v

x = xo + vt v = vo + at x = xo + vot + at2 v2 = vo + 2a(x - xo ) 2 Two types of motion are frequently encountered: uniform rectilinear motion, in which the velocity v of the particle is constant and x = xo + vt and uniformly accelerated rectilinear motion, in which the acceleration a of the particle is constant and v = vo + at 1 2 x = xo + vot + at2 v2 = vo + 2a(x - xo ) 2

xB = xA + xB/A vB = vA + vB/A aB = aA + aB/A O A B x xA xB/A xB When particles A and B move along the same straight line, the relative motion of B with respect to A can be considered. Denoting by xB/A the relative position coordinate of B with respect to A , we have xB = xA + xB/A Differentiating twice with respect to t, we obtain vB = vA + vB/A aB = aA + aB/A where vB/A and aB/A represent, respectively, the relative velocity and the relative acceleration of B with respect to A.

xC xA xB C A B When several blocks are are connected by inextensible cords, it is possible to write a linear relation between their position coordinates. Similar relations can then be written between their velocities and their accelerations and can be used to analyze their motion.

y It is sometimes convenient to resolve the velocity and acceleration of a particle P into components other than the rectangular x, y, and z components. For a particle P moving along a path confined to a plane, we attach to P the unit vectors et tangent to the path and en normal to the path and directed toward the center of curvature of the path. C v 2 r an = en dv dt at = et P x O The velocity and acceleration are expressed in terms of tangential and normal components. The velocity of the particle is v = vet The acceleration is dv dt v2 r a = et + en

v = vet v2 dv a = et + en r dt y C v 2 r an = en dv at = et dt P x O In these equations, v is the speed of the particle and r is the radius of curvature of its path. The velocity vector v is directed along the tangent to the path. The acceleration vector a consists of a component at directed along the tangent to the path and a component an directed toward the center of curvature of the path,

a = (r - rq2)er + (rq + 2rq)eq When the position of a particle moving in a plane is defined by its polar coordinates r and q, it is convenient to use radial and transverse components directed, respectively, along the position vector r of the particle and in the direction obtained by rotating r through 90o counterclockwise. Unit er r = r er P q x O vectors er and eq are attached to P and are directed in the radial and transverse directions. The velocity and acceleration of the particle in terms of radial and transverse components is . . v = rer + rqeq .. . .. . . a = (r - rq2)er + (rq + 2rq)eq

a = (r - rq2)er + (rq + 2rq)eq . . er v = rer + rqeq . .. .. . . r = r er P a = (r - rq2)er + (rq + 2rq)eq q x O In these equations the dots represent differentiation with respect to time. The scalar components of of the velocity and acceleration in the radial and transverse directions are therefore . . vr = r vq = rq . .. .. . . ar = r - rq2 aq = rq + 2rq It is important to note that ar is not equal to the time derivative of vr, and that aq is not equal to the time derivative of vq.

Newton’s Second Law Chapter 12

S Fx = max S Fy = may S Fz = maz To solve a problem involving the motion of a particle, S F = ma should be replaced by equations containing scalar quantities. Using rectangular components of F and a, we have P ax az S Fx = max S Fy = may S Fz = maz x z Using tangential and normal components, y an dv dt S Ft = mat = m at P v2 r S Fn = man = m O x Using radial and transverse components, aq . .. ar S Fr = mar= m(r - rq2) r P .. . q . S Fq = maq = m(rq + 2rq) x O

mv When the only force acting on a particle P is a force F directed toward or away from a fixed point O, the particle is said to be moving under a central force. Since S MO = 0 at any given instant, it follows that HO = 0 for all values of t, and f P mv0 r O f0 r0 P0 . HO = constant We conclude that the angular momentum of a particle moving under a central force is constant, both in magnitude and direction, and that the particle moves in a plane perpendicular to HO .

rmv sin f = romvo sin fo . r2q = h . . mv Recalling that HO = rmv sin f, we have, for points PO and P f P mv0 r rmv sin f = romvo sin fo O f0 r0 P0 for the motion of any particle under a central force. . . Using polar coordinates and recalling that vq = rq and HO = mr2q, we have . r2q = h where h is a constant representing the angular momentum per unit mass Ho/m, of the particle.

Work Energy Chapter 13

Chapter 13 KINETICS OF PARTICLES: ENERGY AND MOMENTUM METHODS Consider a force F acting on a particle A. The work of F corresponding to the small displacement dr is defined as A2 ds s2 A a dr F A1 dU = F dr s s1 Recalling the definition of scalar product of two vectors, dU = F ds cos a where a is the angle between F and dr.

ò ò dU = F dr = F ds cos a U1 2 = F dr U1 2 = (Fxdx + Fydy + Fzdz) s2 The work of F during a finite displacement from A1 to A2 , denoted by U1 2 , is obtained dr F A1 s s1 by integrating along the path described by the particle. A2 ò U1 2 = F dr A1 For a force defined by its rectangular components, we write A2 ò U1 2 = (Fxdx + Fydy + Fzdz) A1

A2 The work of the weight W of a body as its center of gravity moves from an elevation y1 to y2 is obtained by setting Fx = Fz = 0 and Fy = - W . W dy y2 A A1 y y1 y2 ò U1 2 = - Wdy = Wy1 - Wy2 y1 The work is negative when the elevation increases, and positive when the elevation decreases.

ò dU = -Fdx = -kx dx U = - k x dx = kx - kx The work of the force F exerted by a spring on a body A during a finite displacement of the body from A1 (x = x1) to A2 (x = x2) is obtained by writing spring undeformed B A O dU = -Fdx = -kx dx x 2 ò B A 1 U 1 2 = - k x dx x 1 x 1 F 1 2 2 1 2 2 = kx 1 - kx 2 A x The work is positive when the spring is returning to its undeformed position. . B A 2 x 2

The kinetic energy of a particle of mass m moving with a velocity v is defined as the scalar quantity 1 2 T = mv2 From Newton’s second law the principle of work and energy is derived. This principle states that the kinetic energy of a particle at A2 can be obtained by adding to its kinetic energy at A1 the work done during the displacement from A1 to A2 by the force F exerted on the particle: T1 + U1 2 = T2

dU Power = = F v dt h = power output power input The power developed by a machine is defined as the time rate at which work is done: dU dt Power = = F v where F is the force exerted on the particle and v is the velocity of the particle. The mechanical efficiency, denoted by h, is expressed as power output power input h =

U1 2 = V1 - V2 Vg = Wy GMm Vg = - r Ve = kx2 When the work of a force F is independent of the path followed, the force F is said to be a conservative force, and its work is equal to minus the change in the potential energy V associated with F : U1 2 = V1 - V2 The potential energy associated with each force considered earlier is Vg = Wy Force of gravity (weight): GMm r Vg = - Gravitational force: 1 2 Ve = kx2 Elastic force exerted by a spring:

U1 2 = V1 - V2 This relationship between work and potential energy, when combined with the relationship between work and kinetic energy (T1 + U1 2 = T2) results in T1 + V 1 = T2 + V 1 This is the principle of conservation of energy, which states that when a particle moves under the action of conservative forces, the sum of its kinetic and potential energies remains constant. The application of this principle facilitates the solution of problems involving only conservative forces.

Impulse & Momentum Chapter 13

The linear momentum of a particle is defined as the product mv of the mass m of the particle and its velocity v. From Newton’s second law, F = ma, we derive the relation t2 ò mv1 + F dt = mv2 t1 where mv1 and mv2 represent the momentum of the particle at a time t1 and a time t2 , respectively, and where the integral defines the linear impulse of the force F during the corresponding time interval. Therefore, mv1 + Imp1 2 = mv2 which expresses the principle of impulse and momentum for a particle.

mv1 + SImp1 2 = mv2 mv1 + SFDt = mv2 When the particle considered is subjected to several forces, the sum of the impulses of these forces should be used; mv1 + SImp1 2 = mv2 Since vector quantities are involved, it is necessary to consider their x and y components separately. The method of impulse and momentum is effective in the study of impulsive motion of a particle, when very large forces, called impulsive forces, are applied for a very short interval of time Dt, since this method involves impulses FDt of the forces, rather than the forces themselves. Neglecting the impulse of any nonimpulsive force, we write mv1 + SFDt = mv2

In the case of the impulsive motion of several particles, we write Smv1 + SFDt = Smv2 where the second term involves only impulsive, external forces. In the particular case when the sum of the impulses of the external forces is zero, the equation above reduces to Smv1 = Smv2 that is, the total momentum of the particles is conserved.

mAvA + mBvB = mAv’A + mBv’B Line of Impact v’A v’B Before Impact After Impact In the case of direct central impact, two colliding bodies A and B move along the line of impact with velocities vA and vB , respectively. Two equations can be used to determine their velocities v’A and v’B after the impact. The first represents the conservation of the total momentum of the two bodies, mAvA + mBvB = mAv’A + mBv’B

mAvA + mBvB = mAv’A + mBv’B Line of Impact v’A v’B Before Impact After Impact mAvA + mBvB = mAv’A + mBv’B The second equation relates the relative velocities of the two bodies before and after impact, v’B - v’A = e (vA - vB ) The constant e is known as the coefficient of restitution; its value lies between 0 and 1 and depends on the material involved. When e = 0, the impact is termed perfectly plastic; when e = 1 , the impact is termed perfectly elastic.

(vA)t = (v’A)t (vB)t = (v’B)t Line of Impact In the case of oblique central impact, the velocities of the two colliding bodies before and after impact are resolved into n components along the line of impact and t components along the common tangent to the surfaces in contact. In the t direction, n t vB B A Before Impact (vA)t = (v’A)t (vB)t = (v’B)t vA v’B while in the n direction n mA (vA)n + mB (vB)n = mA (v’A)n + mB (v’B)n t v’A B vB (v’B)n - (v’A)n = e [(vA)n - (vB)n] A vA After Impact

(vA)t = (v’A)t (vB)t = (v’B)t Line of Impact n (vA)t = (v’A)t (vB)t = (v’B)t t vB mA (vA)n + mB (vB)n = mA (v’A)n + mB (v’B)n B A Before Impact (v’B)n - (v’A)n = e [(vA)n - (vB)n] vA v’B n Although this method was developed for bodies moving freely before and after impact, it could be extended to the case when one or both of the colliding bodies is constrained in its motion. t v’A B vB A vA After Impact

System of Particle Chapter 14 The theory in this chapter serves as a support to Chapter 16.

Chapter 14 SYSTEMS OF PARTICLES The effective force of a particle Pi of a given system is the product miai of its mass mi and its acceleration ai with respect to a newtonian frame of reference centered at O. The system of the external forces acting on the particles and the system of the effective forces of the particles are equipollent; i.e., both systems have the same resultant and the same moment resultant about O : n n S Fi = S miai i =1 i =1 n n S (ri x Fi ) = S (ri x miai) i =1 i =1

. . S F = L S Mo = Ho L = S mivi Ho = S (ri x mivi) The linear momentum L and the angular momentum Ho about point O are defined as n n L = S mivi Ho = S (ri x mivi) i =1 i =1 It can be shown that . . S F = L S Mo = Ho This expresses that the resultant and the moment resultant about O of the external forces are, respectively, equal to the rates of change of the linear momentum and of the angular momentum about O of the system of particles.

. S F = ma mr = S miri L = mv L = ma . The mass center G of a system of particles is defined by a position vector r which satisfies the equation n mr = S miri i =1 n where m represents the total mass S mi. Differentiating both i =1 members twice with respect to t, we obtain . L = mv L = ma where v and a are the velocity and acceleration of the mass center G. Since S F = L, we obtain . S F = ma Therefore, the mass center of a system of particles moves as if the entire mass of the system and all the external forces were concentrated at that point.

T = S mivi 2 T = mv 2 + S miv’i 2 y’ miv’i The kinetic energy T of a system of particles is defined as the sum of the kinetic energies of the particles. y r’i Pi n G 1 2 T = S mivi x’ 2 O i = 1 x z z’ Using the centroidal reference frame Gx’y’z’ we note that the kinetic energy of the system can also be obtained by adding the kinetic energy mv2 associated with the motion of the mass center G and the kinetic energy of the system in its motion relative to the frame Gx’y’z’ : 1 2 n 1 2 1 2 T = mv 2 + S miv’i 2 i = 1

T = mv 2 + S miv’i 2 T1 + U1 2 = T2 T1 + V1 = T2 + V2 n y’ miv’i 1 1 2 r’i Pi The principle of work and energy can be applied to a system of particles as well as to individual particles G x’ O x z z’ T1 + U1 2 = T2 where U1 2 represents the work of all the forces acting on the particles of the system, internal and external. If all the forces acting on the particles of the system are conservative, the principle of conservation of energy can be applied to the system of particles T1 + V1 = T2 + V2

ò ò (mAvA)1 y y S Fdt y (mBvB)2 (mAvA)2 (mCvC)2 (mBvB)1 O O x x O x S MOdt ò (mCvC)1 t1 The principle of impulse and momentum for a system of particles can be expressed graphically as shown above. The momenta of the particles at time t1 and the impulses of the external forces from t1 to t2 form a system of vectors equipollent to the system of the momenta of the particles at time t2 .

L1 = L2 (HO)1 = (HO)2 (mAvA)1 y y (mBvB)2 (mAvA)2 (mCvC)2 (mBvB)1 O x If no external forces act on the system of particles, the systems of momenta shown above are equipollent and we have L1 = L2 (HO)1 = (HO)2 Many problems involving the motion of systems of particles can be solved by applying simultaneously the principle of impulse and momentum and the principle of conservation of energy or by expressing that the linear momentum, angular momentum, and energy of the system are conserved.

(D m)vB S S F Dt S M Dt B Smivi B Smivi S S A A (D m)vA For variable systems of particles, first consider a steady stream of particles, such as a stream of water diverted by a fixed vane or the flow of air through a jet engine. The principle of impulse and momentum is applied to a system S of particles during a time interval Dt, including particles which enter the system at A during that time interval and those (of the same mass Dm) which leave the system at B. The system formed by the momentum (Dm)vA of the particles entering S in the time Dt and the impulses of the forces exerted on S during that time is equipollent to the momentum (Dm)vB of the particles leaving S in the same time Dt.

dm SF = (vB - vA) dt (D m)vB S F Dt B B Smivi Smivi S S S S M Dt A A (D m)vA Equating the x components, y components, and moments about a fixed point of the vectors involved, we could obtain as many as three equations, which could be solved for the desired unknowns. From this result, we can derive the expression dm dt SF = (vB - vA) where vB - vA represents the difference between the vectors vB and vA and where dm/dt is the mass rate of flow of the stream.

Kinematics of Rigid Bodies Chapter 15

Chapter 15 KINEMATICS OF RIGID BODIES z In rigid body translation, all points of the body have the same velocity and the same acceleration at any given instant. A’ B q P f Considering the rotation of a rigid body about a fixed axis, the position of the body is defined by the angle q that the line BP, drawn from the axis of rotation to a point P of the body, r O x y A forms with a fixed plane. The magnitude of the velocity of P is . ds dt v = = rq sin f . where q is the time derivative of q.

. ds v = = rq sin f dt dr v = = w x r dt . w = wk = qk z A’ The velocity of P is expressed as B q dr dt P f v = = w x r r O x where the vector y A . w = wk = qk is directed along the fixed axis of rotation and represents the angular velocity of the body.

. dr v = = w x r dt w = wk = qk a = a x r + w x (w x r) .. . z dr dt . A’ v = = w x r w = wk = qk B Denoting by a the derivative dw/dt of the angular velocity, we express the acceleration of P as q P f r O x a = a x r + w x (w x r) y A differentiating w and recalling that k is constant in magnitude and direction, we find that . .. a = ak = wk = qk The vector a represents the angular acceleration of the body and is directed along the fixed axis of rotation.

v = wk x r at = ak x r at = ra an= -w2 r an = rw2 y Consider the motion of a representative slab located in a plane perpendicualr to the axis of rotation of the body. The angular velocity is perpendicular to the slab, so the velocity of point P of the slab is v = wk x r P O r x w = wk v = wk x r where v is contained in the plane of the slab. The acceleration of point P can be resolved into tangential and normal components respectively equal to y at = ak x r P an= -w2 r O x w = wk at = ak x r at = ra a = ak an= -w2 r an = rw2

dq w = dt dw d2q a = = dt dt2 dw w = a dq The angular velocity and angular acceleration of the slab can be expressed as dq dt w = dw dt d2q dt2 a = = or dw dq w = a Two particular cases of rotation are frequently encountered: uniform rotation and uniformly accelerated rotation. Problems involving either of these motions can be solved by using equations similar to those for uniform rectilinear motion and uniformly accelerated rectilinear motion of a particle, where x, v, and a are replaced by q, w, and a.

vB = vA + vB/A vA vA wk vB vA vB/A y’ wk vB (fixed) vA x’ A A A rB/A vB/A B B B Plane motion = Translation with A + Rotation about A The most general plane motion of a rigid slab can be considered as the sum of a translation and a rotation. The slab shown can be assumed to translate with point A, while simultaneously rotating about A. It follows that the velocity of any point B of the slab can be expressed as vB = vA + vB/A where vA is the velocity of A and vB/A is the relative velocity of B with respect to A.

vB/A = wk x rB/A vB/A = (rB/A )w = rw vA vA vB/A y’ wk vB (fixed) vA x’ vA vB A A A rB/A vB/A B B B Plane motion = Translation with A + Rotation about A vB = vA + vB/A Denoting by rB/A the position of B relative to A, we note that vB/A = wk x rB/A vB/A = (rB/A )w = rw The fundamental equation relating the absolute velocities of points A and B and the relative velocity of B with respect to A can be expressed in the form of a vector diagram and used to solve problems involving the motion of various types of mechanisms.

Another approach to the solution of problems involving the velocities of the points of a rigid slab in plane motion is based on determination of the instantaneous center of rotation C of the slab. C C vB B vB A vA vA

aB = aA + aB/A aA aA aB/A aB (aB/A)n aA (aB/A)t y’ aA wk aA ak aB/A aB x’ A A A (aB/A)n aA (aB/A)t B B B Plane motion = Translation with A + Rotation about A The fact that any plane motion of a rigid slab can be considered the sum of a translation of the slab with reference to point A and a rotation about A is used to relate the absolute accelerations of any two points A and B of the slab and the relative acceleration of B with respect to A. aB = aA + aB/A where aB/A consists of a normal component (aB/A )n of magnitude rw2 directed toward A, and a tangential component (aB/A )t of magnitude ra perpendicular to the line AB.

aB = aA + aB/A aA aA aB/A aB (aB/A)t (aB/A)n aA y’ aA wk aA ak aB/A aB x’ A A A (aB/A)t (aB/A)n aA B B B Plane motion = Translation with A + Rotation about A aB = aA + aB/A The fundamental equation relating the absolute accelerations of points A and B and the relative acceleration of B with respect to A can be expressed in the form of a vector diagram and used to determine the accelerations of given points of various mechanisms. (aB/A)n (aB/A)t aA aB aB/A

aB = aA + aB/A aA aA aB/A aB (aB/A)t (aB/A)n aA y’ aA wk aA ak aB/A aB x’ A A A (aB/A)t (aB/A)n aA B B B Plane motion = Translation with A + Rotation about A aB = aA + aB/A (aB/A)n (aB/A)t aA aB aB/A The instantaneous center of rotation C cannot be used for the determination of accelerations, since point C , in general, does not have zero acceleration.

Inertia Chapter 9 It is unlikely that you will calculate the mass moment of inertia. But the theory behind it has to be covered.

Chapter 9 DISTRIBUTED FORCES: MOMENTS OF INERTIA y The rectangular moments of inertia Ix and Iy of an area are defined as y x Ix = y 2dA Iy = x 2dA ò ò dx x These computations are reduced to single integrations by choosing dA to be a thin strip parallel to one of the coordinate axes. The result is 1 3 dIx = y 3dx dIy = x 2ydx

y The polar moment of inertia of an area A with respect to the pole O is defined as dA r y JO = r 2dA ò O x x A The distance from O to the element of area dA is r. Observing that r 2 =x 2 + y 2 , we established the relation JO = Ix + Iy

y The radius of gyration of an area A with respect to the x axis is defined as the distance kx, where Ix = kx A. With similar definitions for the radii of gyration of A with respect to the y axis and with respect to O, we have A kx 2 O x Ix A Iy A JO A kx = ky = kO =

The parallel-axis theorem states that the moment of inertia I of an area with respect to any given axis AA’ is equal to the moment of inertia I of the area with respect to the centroidal c B B’ d A A’ axis BB’ that is parallel to AA’ plus the product of the area A and the square of the distance d between the two axes: I = I + Ad 2 This expression can also be used to determine I when the moment of inertia with respect to AA’ is known: I = I - Ad 2

A similar theorem can be used with the polar moment of inertia A similar theorem can be used with the polar moment of inertia. The polar moment of inertia JO of an area about O and the polar moment of inertia JC of the area about its c d o centroid are related to the distance d between points C and O by the relationship JO = JC + Ad 2 The parallel-axis theorem is used very effectively to compute the moment of inertia of a composite area with respect to a given axis.

ò I = r 2dm I m k = A’ Moments of inertia of mass are encountered in dynamics. They involve the rotation of a rigid body about an axis. The mass moment of inertia of a body with respect to an axis AA’ is defined as r1 Dm1 Dm3 r2 r3 Dm2 I = r 2dm ò A where r is the distance from AA’ to the element of mass. The radius of gyration of the body is defined as I m k =

ò ò ò Ix = (y 2 + z 2 ) dm Iy = (z 2 + x 2 ) dm Iz = (x 2 + y 2 ) dm The moments of inertia of mass with respect to the coordinate axes are Ix = (y 2 + z 2 ) dm ò Iy = (z 2 + x 2 ) dm ò Iz = (x 2 + y 2 ) dm ò A’ d The parallel-axis theorem also applies to mass moments of inertia. B’ I = I + d 2m A G I is the mass moment of inertia with respect to the centroidal BB’ axis, which is parallel to the AA’ axis. The mass of the body is m. B

The moments of inertia of thin plates can be readily obtained from the moments of inertia of their areas. For a rectangular plate, the moments of inertia are 1 12 1 12 IAA’ = ma 2 IBB’ = mb 2 1 12 ICC’ = IAA’ + IBB’ = m (a 2 + b 2) A’ B’ For a circular plate they are r t C IAA’ = IBB’ = mr 2 1 4 C’ ICC’ = IAA’ + IBB’ = mr 2 1 2 B A

PLANE MOTION OF RIGID BODIES: FORCES AND ACCELERATIONS Chapter 16

Chapter 16 PLANE MOTION OF RIGID BODIES: FORCES AND ACCELERATIONS . HG The relations existing between the forces acting on a rigid body, the shape and mass of the body, and the motion produced are studied as the kinetics of rigid bodies. In general, our analysis is restricted to the plane motion of F4 F1 ma F3 G G F2 rigid slabs and rigid bodies symmetrical with respect to the reference plane.

. The two equations for the motion of a system of particles apply to the most general case of the motion of a rigid body. The first equation defines the motion of the mass center G of the body. HG F4 F1 ma F3 G G F2 SF = ma where m is the mass of the body, and a the acceleration of G. The second is related to the motion of the body relative to a centroidal frame of reference. . SMG = HG

. HG SF = ma F4 . F1 . ma SMG = HG . where HG is the rate of change of the angular momentum HG of the body about its mass center G. F3 G G F2 These equations express that the system of the external forces is equipollent to the system consisting of the vector ma attached at G and the couple of moment HG. .

. HG For the plane motion of rigid slabs and rigid bodies symmetrical with respect to the reference plane, the angular momentum of the body is expressed as F4 F1 ma F3 G G F2 HG = Iw where I is the moment of inertia of the body about a centroidal axis perpendicular to the reference plane and w is the angular velocity of the body. Differentiating both members of this equation . . HG = Iw = Ia

SFx = max SFy = may SMG = Ia For the restricted case considered here, the rate of change of the angular momentum of the rigid body can be represented by a vector of the same direction as a (i.e. ma F3 G G Ia F2 perpendicular to the plane of reference) and of magnitude Ia. The plane motion of a rigid body symmetrical with respect to the reference plane is defined by the three scalar equations SFx = max SFy = may SMG = Ia The external forces acting on a rigid body are actually equivalent to the effective forces of the various particles forming the body. This statement is known as d’Alembert’s principle.

F4 F1 d’Alembert’s principle can be expressed in the form of a vector diagram, where the effective forces are represented by a vector ma attached at G and a couple Ia. In the case of a slab in translation, the effective forces (part b of the figure) reduce to a ma F3 G G Ia F2 (a) (b) single vector ma ; while in the particular case of a slab in centroidal rotation, they reduce to the single couple Ia ; in any other case of plane motion, both the vector ma and Ia should be included.

F4 Any problem involving the plane motion of a rigid slab may be solved by drawing a free-body-diagram equation similar to that shown. Three equations of of motion can then be obtained by equating the x components, F1 ma F3 G G Ia F2 y components, and moments about an arbitrary point A, of the forces and vectors involved. This method can be used to solve problems involving the plane motion of several connected rigid bodies. Some problems, such as noncentroidal rotation of rods and plates, the rolling motion of spheres and wheels, and the plane motion of various types of linkages, which move under constraints, must be supplemented by kinematic analysis.

PLANE MOTION OF RIGID BODIES: ENERGY AND MOMENTUM METHODS Chapter 17

Chapter 17 PLANE MOTION OF RIGID BODIES: ENERGY AND MOMENTUM METHODS The principle of work and energy for a rigid body is expressed in the form T1 + U1 2 = T2 where T1 and T2 represent the initial and final values of the kinetic energy of the rigid body and U1 2 the work of the external forces acting on the rigid body. The work of a force F applied at a point A is s2 ò U1 2 = (F cos a) ds s1 where F is the magnitude of the force, a the angle it forms with the direction of motion of A, and s the variable of integration measuring the distance traveled by A along its path.

The work of a couple of moment M applied to a rigid body during a rotation in q of the rigid body is ò U1 2 = M ds q1 The kinetic energy of a rigid body in plane motion is 1 2 1 2 T = mv 2 + Iw2 G w v where v is the velocity of the mass center G of the body, w the angular velocity of the body, and I its moment of inertia about an axis through G perpendicular to the plane of reference.

1 2 1 2 T = mv 2 + Iw2 G The kinetic energy of a rigid body in plane motion may be separated into two parts: (1) the kinetic energy mv 2 associated with the motion of the mass center G of the w v 1 2 1 2 body, and (2) the kinetic energy Iw2 associated with the rotation of the body about G. For a rigid body rotating about a fixed axis through O with an angular velocity w, O 1 2 T = IOw2 w where IO is the moment of inertia of the body about the fixed axis.

T1 + V1 = T2 + V2 dU dt M dq dt Power = = = Mw When a rigid body, or a system of rigid bodies, moves under the action of conservative forces, the principle of work and energy may be expressed in the form which is referred to as the principle of conservation of energy. This principle may be used to solve problems involving conservative forces such as the force of gravity or the force exerted by a spring. T1 + V1 = T2 + V2 The concept of power is extended to a rotating body subjected to a couple dU dt M dq dt Power = = = Mw where M is the magnitude of the couple and w is the angular velocity of the body.

Syst Momenta1 + Syst Ext Imp1 2 = Syst Momenta2 The principle of impulse and momentum derived for a system of particles can be applied to the motion of a rigid body. Syst Momenta1 + Syst Ext Imp1 2 = Syst Momenta2 For a rigid slab or a rigid body symmetrical with respect to the reference plane, the system of the momenta of the particles forming the body is equivalent to a vector mv attached to the mass center G of the body and a couple Iw. The vector mv is associated with translation of the body with G and represents the linear momentum of the body, while the couple Iw corresponds to the rotation of the body about G and represents the angular momentum of the body about an axis through G. (Dm)v mv P Iw

The principle of impulse and momentum can be expressed graphically by drawing three diagrams representing respectively the system of initial momenta of the body, the impulses of the external forces acting on it, and the system of the final momenta of the body. Summing and equating respectively the x components, the y components, and the moments about any given point of the vectors shown in the figure, we obtain three equations of motion which may be solved for the desired unknowns. ò Fdt mv2 y mv1 y y G G Iw1 Iw2 O x O x O x

ò Fdt mv2 y mv1 y y G G Iw1 Iw2 O x O x O x In problems dealing with several connected rigid bodies each body may be considered separately or, if no more than three unknowns are involved, the principles of impulse and momentum may be applied to the entire system, considering the impulses of the external forces only. When the lines of action of all the external forces acting on a system of rigid bodies pass through a given point O, the angular momentum of the system about O is conserved.

(v’B)n - (v’A)n = e[(vA)n - (vB)n] The eccentric impact of two rigid bodies is defined as an impact in which the mass centers of the colliding bodies are not located on the line of impact. In such a situation a relation for the impact involving the coefficient of restitution e holds, and the velocities of points A and B where contact occurs during the impact should be used. B A vB n vA (a) Before impact n B A v’B (v’B)n - (v’A)n = e[(vA)n - (vB)n] n v’A (b) After impact

(v’B)n - (v’A)n = e[(vA)n - (vB)n] (a) Before impact (b) After impact (v’B)n - (v’A)n = e[(vA)n - (vB)n] where (vA)n and (vB)n are the components along the line of impact of the velocities of A and B before impact, and (v’A)n and (v’B)n their components after impact. This equation is applicable not only when the colliding bodies move freely after impact but also when the bodies are partially constrained in their motion.