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Chapter 3 – KINEMATICS OF RIGID BODIES Serge Abrate Department of Mechanical Engineering and Energy Processes Southern Illinois University abrate@engr.siu.edu After studying the dynamics of particles we now turn to the study of rigid bodies. A rigid body is defined as a set of particles in which the distance between any two points remains constant. This B distinguishes rigid bodies from fluids and granular materials such as sand for example. The distinction between rigid bodies and deformable bodies is based on the type of analysis that is performed. All solids are deformable but, in the A types of problems covered in dynamics, the deformations of the body are negligible compared to the overall motion. Figure 1: In a rigid body the distance between any two points A and B remains 1-TWO-DIMENSIONAL MOTION constant Considering the motion of a lamina in the xy plane, we examine two simple types of motion (translation and rotation) and show that the general motion consists of a combination of the two. Then V we introduce the concept of the instantaneous center of rotation which is useful in many practical problems. V B V 1.1. Translation C A Translation is a type of motion in which at a given instant every point in the body has the same velocity V . That is, the velocity has the same direction and the same magnitude at every point. Figure 2: During translation, at That does not necessarily mean that the body each instant, the velocity vector moves on a straight line since V can change with is the same for all the points time V V t . 1.2. Rotation about a fixed point 1.2.1- Velocity of a point in a body rotating about a fixed point For a body rotating about a fixed point O (Fig. 3), the trajectory of an arbitrary point A is a circle of radius rA . In the radial and tangential coordinate system shown in Fig. 4, the position of A is given by rA r er (1) Taking the derivative with respect to time, the velocity of A is y drA dr de vA er r r (2) dt dt dt A rA Since the body is rigid, the radius r does not change dr and 0 . Using the chain rule of differentiation, dt O x d er d er d (3) Figure 3: Rotation about a dt d dt fixed point O As the angle increases by an amount d , y et the unit vector e r becomes e r and d er er er d et . Then, er A d er et (4) d rA Substituting into Eq. gives O x v A r et (5) Generally denotes the angular velocity so the velocity of A can also be written as Figure 4 : Radial and tangential coordinates v A r et (6) As the angle increases by an amount d , point A travels a distance rd . If the magnitude of the velocity is v, the distance traveled is also equal to vdt. The equality rd vdt gives the following relationship between the linear velocity at A and the angular velocity of the body d vr r (7) dt so v A v et (8) Eqs. 5,6 and 8 indicate that in the case of a rotation about a fixed point, the velocity is always oriented in the tangential direction and that its magnitude is increases linearly with the radius r and the angular velocity. 1.2.2- Acceleration of a point in a body rotating about a fixed point To find the acceleration at point A, take the derivative with respect to time of the velocity given by Eq. 6 d vA d de aA r et r t (9) dt dt dt d is the angular acceleration which is usually called and dt d et d et d er (10) dt d dt Then a A r et r2 er (11) 1.2.3- Velocity and acceleration in vector form Here we introduce the angular velocity vector which is perpendicular to the plane of motion and has a magnitude . is then in the z direction (Fig. 3) and is positive for a rotation in the counter-clockwise direction. will be in the negative direction if the body rotates clockwise. Then we notice that , rA and v A form a right handed system and that the magnitude of the velocity is the product of the magnitude of the two other vectors. Therefore, v A x rA (12) The acceleration is found by taking the derivative with respect to time and d v A d dr aA x rA x A x rA x v A (13) dt dt dt where we have introduced the angular acceleration vector x rA . Using Eq. 12, we obtain a A x rA x x rA (14) Just as v A x rA is acting in the tangential direction, so is x rA . Using the following identity for the triple cross product We find that x x rA . rA . rA . Since the angular velocity is normal to the plane of motion, the first term on the rhs is zero. Then, the acceleration in Eq. 14 is seen to contain one term in the tangential direction and another in the radial direction and in fact Eq. 14 is identical to Eq. 11. 1.3- General two-dimensional motion Consider two points A and B in a body moving in the xy plane (Figure 5). Since y B rB rA rB / A (15) A rB / A the velocity of B is v B v A v B / A (16) rA x Since the body is rigid, the relative velocity v B / A must be a Figure 5 : Two- rotation about A in the xy-plane. That is, dimensional motion v B / A x rB / A (17) Substituting into Eq. 16 gives v B v A x rB / A (18) Therefore, the motion of the body consists of a translation with a velocity v A , the velocity of the reference point A, plus a rotation about point A. The acceleration of point B is v B v A x rB / A x rB / A d d d d aB (19) dt dt dt dt or a B a A x rB / A x v B / A (20) Finally, a B a A x rB / A x x rB / A (21) Eqs. 18, 21 describe the velocity and acceleration of a lamina in general 2D motion. Later we will see that the same equations apply for the 3D case. 1.4- Invariance of the angular velocity and acceleration vectors The previous section showed that motion of an arbitrary point B can be written in terms of the motion of another arbitrary point A, an angular velocity and an angular acceleration . This brings up the question: Are those vectors and unique or do they depend on the choice of point A? To address this question, we write the velocity of B in terms of the velocity of another arbitrary point C as v B v C x rB / C (22) where we allow for the possibility that is different than . We also express the velocity of C in terms of the velocity of A v C v A x rC / A (23) Substituting Eq. 23 into Eq. 22 gives v B v A x rC / A x rB / C (24) Recalling Eq. 18 we find that v A x rC / A x rB / C v A x rB / A (25) or x rB / C x rB / A rC / A (26) Since rB / A rC / A rB / C , we conclude that . In other words, the angular velocity is the same no matter which point in the body is taken as the reference. This is why it is called the angular velocity of the body. Since , their derivatives, the angular accelerations are also the same. 1.5- Instantaneous center of rotation In two-dimensional the instantaneous center of rotation (I.C.) concept is used to solve problems most efficiently. The main idea is that at each instant there is a point with zero velocity and the body simply rotates about that point. Of course as time goes on, the location of the I.C. changes. 1.5.1- Existence of the instantaneous center of rotation Given the velocity of an arbitrary point A and the angular velocity of the body we seek to determine the location of a point O where the velocity is zero at that instant. Using Eq. 18, v O v A x rO / A 0 (27) The cross product v A x v O can be written as v A x v A v A x x rO / A 0 (28) The first term being zero, the triple cross product is expanded so that Eq. 28 becomes v A . rO / A v A . rO / A 0 (29) Because is perpendicular to the plane of motion, the second term in Eq. 29 is zero and then, v A . rO / A . Therefore, point O is located on a line that is perpendicular to the velocity vector v A . Taking the cross product of and v O gives x v O x v A x x rO / A 0 (29) Using Lagrange’s formula for the triple product we find that x v A . rO / A . rO / A 0 (30) Examining each term in this equation we find that, by definition, x v A 0 since we have 2D motion and is perpendicular to the plane of the motion. The second term . 2 . Then, Eq. 30 becomes rO / A 2 x v A 1 (31) If both and v A are known, Eq. 31 provides a convenient way for determining the location of point O. This equation shows that such a point O exists provided that 0 (otherwise there is no rotation) and that it is located on a line that is perpendicular to the velocity vector and that , v A and rO / A form a right handed system. O Using Eq. 18, the velocity of an arbitrary point B can I.C. be written as v B v A x rB / A and since vB v A x rA / O , v B x rA / O x rB / A x rB / O . Therefore, the line OB is perpendicular to the velocity vector at B and the body, at that instant, is simply rotating about O. This is why it is called the instantaneous center of rotation. B 1.5.2- Graphical construction A vA Given the velocity vectors at two points A and B (Fig. 7), draw a line normal to at A and a line normal to at B. These two lines intersect at the I.C. Then, the angular velocity can be calculated from Figure 7 : Finding the location of the I.C. graphically v A OA . or v B OB . (32) Since there can be only one angular velocity for a rigid body, vA v B (33) OA OB which leads to a consistency requirement between the geometry and the velocities v A OA (34) v B OB For a given geometry, we can find the I.C. using this graphical construction but the specified velocities must satisfy Eq. (34). 1.5.3- Special case I: the velocity at point A is the same as the velocity at B In many problems is unknown but the velocity is known at two points A and B. Using Eq. 18 we write v O v A x rO / A 0 and v O v B x rO / B 0 (35) Subtracting those two equations gives v A v B x rO / A rO / B 0 or v A v B x rA / O rB / O 0 (36) or v A v B x rA / B (37) Eq. 37 shows that, when two distinct points A and B have the same velocity, the angular velocity is zero and the body is in pure translation. 1.5.4- Special case II: the velocity at A is parallel to the velocity at B Assume that v A is parallel to v B but that the magnitudes are different. What can we conclude? Since these two vectors are parallel, v Ax v B 0 (38) From Eq. 18 we can write v B v A x rB / A (39) Substituting into Eq. 38 gives v A x v A v A x x rB / A 0 . The first term is zero and expanding the triple cross-product gives v A . rB / A v A . rB / A 0 (40) v A . 0 for 2D motion and Eq. 40 becomes v A . rB / A 0 (41) Eq. 41 indicates that there are two cases. With case I, the velocities are perpendicular to the line AB so v A . rB / A 0 and the instantaneous center is located on AB. When v B v A the I.C. will be located at a finite distance from A and B and the angular velocity will have a finite value (Fig. 6.a). When v B v A the I.C. will be located at infinity and the angular velocity will be zero (Fig. 6.b). This is a case of pure translation. With case II, the velocities are not perpendicular to AB: v A . rB / A 0 . In that case, Eq. 41 implies that 0 which suggests that the motion should be a pure translation. However since v B v A this situation is impossible (Fig. 6.c). I.C. I.C. vA vA A A vA vB vB vB B B B (a) (b) (c) Figure 6 : (a) velocities at A and B are parallel and normal to AB and have different magnitudes; (b) velocities at A and B are the same and are normal to AB (translation) ; (c) velocities at A and B are parallel with different magnitudes and are not normal to AB (impossible case) 2- GENERAL MOTION OF A RIGID BODY z In order to describe the motion of a rigid body in three dimensions, we use a fixed coordinate system XYZ and another coordinate system O-xyz that is fixed to y the body (Fig. 7). Point O is an arbitrary point in the body. The angular velocity and the angular acceleration are two vectors that can change O P orientation and magnitude with time. rP / O Z rO The position of P, an arbitrary point in the body, is defined by the position vector x Y rP rO rP / O (42) X The velocity of that point P is Figure 7 : Position vector and local drP drO drP / O coordinate system vP (43) dt dt dt The first term is simply the velocity of the origin of the moving frame drO vO (44) dt In the moving coordinate system, if rP / O x i y j z k then d rP / O di dj dk xi y j zk x y z (45) dt dt dt dt Since the body is rigid, x y z 0 and because the moving frame is rotating with an angular velocity , di dj dk x i, x j, x k (46) dt dt dt The velocity of P relative to O becomes vP / O d rP / O dt x x i y j z k x rP / O (47) So, v P v O x rP / O (48) The acceleration of P is found by taking the derivative with respect to time so d v P dv O d dr aP x rP / O x P / O (49) dt dt dt dt or a P a O x rP / O x x rP / O (50) d where the angular acceleration . dt Note that the expressions for the velocity and the acceleration of an arbitrary point in terms of a reference point and the angular velocity and acceleration of the body are identical when written in vector form. 3- MOTION OF A PARTICLE RELATIVE TO A RIGID BODY As before, we consider the motion of a rigid body z and the xyz coordinate system is fixed to the body and rotates with an angular velocity and an angular acceleration . P is an arbitrary point in y the body (Figure 8). O P In addition, there is a particle located at point P’. rP / O The position of P’ is given by Z rO P’ rP ' ro rP / O rP '/ P (51) x Y Taking the derivative with respect to time, the velocity of P’ is X v P ' v o x rP / O x rP '/ P v P '/ xy z (52) Figure 8 : Position vector and local coordinate system Taking the derivative with respect to time again gives the acceleration of P’ d dr d dr dv P '/ xy z a P' a o x rP / O x P / O x rP '/ P x P '/ P x v P '/ xy z (53) dt dt dt dt dt a P ' a o x rP / O x x rP / O x rP '/ P x v P '/ xy z x v P '/ xy z a P '/ xy z (54) Equations 52 and 54 simplify a little bit when P’ is at P rP '/ P 0 . In that case, the velocity and acceleration of the moving particle are v P ' v o x rP / O v P '/ xy z (55) a P ' a o x rP / O x x rP / O 2 x v P '/ xy z a P '/ xy z (56) Equation 55 says that the velocity of the moving particle is equal the velocity of the point on the rigid body that occupies the same location at that instant v o x rP / O plus the velocity of the particle relative to the body v P '/ xy z . In Equation 56, we have the acceleration of point P a o x rP / O x x rP / O , the acceleration of P’ relative to the body a P '/ xy z , and then an additional term 2 x v P '/ xy z which is called Coriolis’ acceleration. It is named after Gaspard- Gustave Coriolis (1792-1843), a French scientist who in 1835, observed an apparent deflection of moving objects from a straight path when they are viewed from a rotating frame. Appendix: Vector algebra Cross product The dot product of two vectors U U x i U y j Uz k and V Vx i Vy j Vz k can be written in terms of the components of the two vectors as U . V Ux Vx U y Vy Uz Vz or in terms of the magnitude of the two vectors and the angle between them as U . V U V cos Then it becomes obvious that when the two vectors U and V are orthogonal, U . V 0 . Cross product The cross product of two vectors U and V is a vector W of magnitude W U V sin . U, V and W form a right angled system. The components of W can be determined from the determinant i j k W U x V Ux Uy U z i U y Vz U z Vy j U x Vz U z Vx k U x Vy U y Vx Vx Vy Vz A straightforward expansion will show that U x V V x U . Similarly, U x c U 0 so, the dot product of two vectors that are parallel is zero. - Angular velocity and acceleration - Vector notation Rotation about v A x rA Translation a fixed point a A x rA x x rA v B v A x rB / A a B a A x rB / A x x rB / A Two-dimensional motion I.C. of rotation 3D motion v P v O x rP / O a P a O x rP / O x x rP / O Figure 8 : Roadmap for this chapter Triple product The cross product of a vector with a cross product is called a triple cross product. The expansion formula of the triple cross product (Lagrange’s formula) is a x b x c a . c b a . b c

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