Chapter 16: Rigid Body Motion

Section 16.1 Rigid Body Motion Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid Body Motion There are cases where an object cannot be treated as a particle. In these cases the size or shape of the body must be considered. Rotation of the body about its center of mass requires a different approach. For example, in the design of gears, cams, and links in machinery or mechanisms, rotation of the body is an important aspect in the analysis of motion. We will now start to study rigid body motion. The analysis will be limited to planar motion. A body is said to undergo planar motion when all parts of the body move along paths equidistant from a fixed plane. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Planar Rigid Body Motion (1 of 4) There are three types of planar rigid body motion. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Planar Rigid Body Motion (2 of 4) Translation: Translation occurs if every line segment on the body remains parallel to its original direction during the motion. When all points move along straight lines, the motion is called rectilinear translation. When the paths of motion are curved lines, the motion is called curvilinear translation. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Planar Rigid Body Motion (3 of 4) Rotation about a fixed axis: In this case, all the particles of the body, except those on the axis of rotation, move along circular paths in planes perpendicular to the axis of rotation. General plane motion: In this case, the body undergoes both translation and rotation. Translation occurs within a plane and rotation occurs about an axis perpendicular to this plane. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Planar Rigid Body Motion (4 of 4) An example of bodies undergoing the three types of motion is shown in this mechanism. The wheel and crank undergo rotation about a fixed axis. In this case, both axes of rotation are at the location of the pins and perpendicular to the plane of the figure. The piston undergoes rectilinear translation since it is constrained to slide in a straight line. The connecting rod undergoes curvilinear translation, since it will remain horizontal as it moves along a circular path. The connecting rod undergoes general plane motion, as it will both translate and rotate. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid-body Motion: Translation The positions of two points A and B on a translating body can be related by rB  rA  rB / A where rA & rB are the absolute position vectors defined from the fixed x-y coordinate system, and rB / A is the relative-position vector between B and A. vB  v A  drB / A / dt drB / A / dt  0 since rB / A is constant. So, vB  v A , The velocity at B is Now and by following similar logic, aB  a A . Note, all points in a rigid body subjected to translation move with the same velocity and acceleration. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Section 16.3 Rigid-body Motion: Rotation About A Fixed Axis Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid-body Motion: Rotation About A Fixed Axis (1 of 2) When a body rotates about a fixed axis, any point P in the body travels along a circular path. The angular position of P is defined by  The change in angular position, d, is called the angular displacement, with units of either radians or revolutions. They are related by 1 revolution  2  radians Angular velocity, , is obtained by taking the time derivative of angular displacement: Similarly, angular acceleration is Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid-body Motion: Rotation About A Fixed Axis (2 of 2) If the angular acceleration of the body is constant,    C the equations for angular velocity and acceleration can be integrated to yield the set of algebraic equations below.   0   C t   0  0t  0.5 C t 2 2  0   2 C   0  2 0 and 0 are the initial values of the body’s angular position and angular velocity. Note these equations are very similar to the constant acceleration relations developed for the rectilinear motion of a particle. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid-body Rotation: Velocity of Point P (1 of 2) The magnitude of the velocity of P is equal to r (the text provides the derivation). The velocity’s direction is tangent to the circular path of P. In the vector formulation, the magnitude and direction of v can be determined from the cross product of  and rp . here rp is a vector from any point on the axis of rotation to P. v  w  rp  w  r The direction of v is determined by the right-hand rule. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid-body Rotation: Acceleration of Point P (1 of 2) The acceleration of P is expressed in terms of its normal an  and tangential at  2 a   r and a   r. In scalar form, these are t n The tangential component, at  represents the time rate of change in the velocity's magnitude. It is directed tangent to the path of motion. The normal component, an , represents the time rate of change in the velocity’s direction. It is directed toward the center of the circular path. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rigid-body Rotation: Acceleration of Point P (2 of 2) Using the vector formulation, the acceleration of P can also be defined by differentiating the velocity. a  dv / dt  dw / dt  rp    drp / dt    rp      rp  It can be shown that this equation reduces to a    r  2 r  at  an The magnitude of the acceleration vector is a an   at  2 2 Copyright © 2016 Pearson Education, Inc. All Rights Reserved Rotation About A Fixed Axis : Procedure • Establish a sign convention along the axis of rotation. • If a relationship is known between any two of the variables (,,, or t), the other variables can be determined from the equations:   d  / dt   d  / dt  d    d  • If  is constant, use the equations for constant angular acceleration. • To determine the motion of a point, the scalar equations 2 2 2 V  r , at  r , an   r and a  an   at  can be used. • Alternatively, the vector form of the equations can be used (with i, j, k components). v    rp    r a  at  an    rp    rp     r  2 r Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis When a body is subjected to general plane motion, it undergoes a combination of translation and rotation. Point A is called the base point in this analysis. It generally has a known motion. The x’- y’ frame translates with the body, but does not rotate. The displacement of point B can be written: Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis: Velocity (3 of 5) The velocity at B is given as : (drB / dt)  (drA / dt )  (drB νB  ν A  νB A A / dt) or Since the body is taken as rotating about A, νB A  drB A / dt  ω  rB A Here ω will only have a k component since the axis of rotation is perpendicular to the plane of translation. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis: Velocity (4 of 5) νB  ν A  ω  rB A When using the relative velocity equation, points A and B should generally be points on the body with a known motion. Often these points are pin connections in linkages. For example, point A on link AB must move along a horizontal path, whereas point B moves on a circular path. The directions of ν A and νB are known since they are always tangent to their paths of motion. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis: Velocity (5 of 5) νB  ν A  ω  rB A When a wheel rolls without slipping, point A is often selected to be at the point of contact with the ground. Since there is no slipping, point A has zero velocity. Furthermore, point B at the center of the wheel moves along a horizontal path. Thus, νB has a known direction, e.g., parallel to the surface. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 2 (1 of 2) Given: Crank rotates OA with an angular velocity of 12 rad/s Find: The velocity of piston B and the angular velocity of rod AB. Plan: Notice that point A moves on a circular path. The directions of ν A is tangent to its path of motion. Draw A kinematic diagram of rod AB and use νB  ν A  ωAB  rB/A. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example 2 (2 of 2) Solution: Since crack OA rotates with an angular velocity of 12 rad/s, the velocity at A will be: ν A  0.3( 12 )i  3.6i m/s Rod AB. Write the relative-velocity equation: νB  ν A  ωAB  rB/A.  νB j  3.6i  ωAB k  ( 0.6 cos 30i  0.6 sin 30 j)] νB j  3.6i  0.5196ωAB j  0.3ωAB i By comparing the i, j components: i: 0  3.6  0.3 ωAB  ωAB  12 rad/s j: νB  0.5196 ωAB  νB  6.24 m/s Copyright © 2016 Pearson Education, Inc. All Rights Reserved Applications (2 of 3) The large window is opened using a hydraulic cylinder AB. The position B of the hydraulic cylinder rod is related to the angular position, q, of the window. A designer has to relate the translational velocity at B of the hydraulic cylinder and the angular velocity and acceleration of the window? How would you go about the task? Copyright © 2016 Pearson Education, Inc. All Rights Reserved Applications (3 of 3) The position of the piston, x, can be defined as a function of the angular position of the crank, q. By differentiating x with respect to time, the velocity of the piston can be related to the angular velocity, w, of the crank. This is necessary when designing an engine. The stroke of the piston is defined as the total distance moved by the piston as the crank angle varies from 0 to 180°. How does the length of crank AB affect the stroke? Copyright © 2016 Pearson Education, Inc. All Rights Reserved Absolute Motion Analysis (Section 16.4) The figure below shows the window using a hydraulic cylinder AB. The absolute motion analysis method relates the position of a point, B, on a rigid body undergoing rectilinear motion to the angular position, q, of a line contained in the body. Once a relationship in the form of sB = f  q  is established, the velocity and acceleration of point B are obtained in terms of the angular velocity and angular acceleration of the rigid body by taking the first and second time derivatives of the position function. Usually the chain rule must be used when taking the derivatives of the position coordinate equation. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Procedure for Absolute Motion Analysis The velocity and acceleration of a point undergoing rectilinear motion can be related to the angular velocity and angular acceleration of a line contained within a body using the following procedure. 1. Locate point on the body using position coordinate s, which is measured from a fixed origin. 2. From a fixed reference line, measure the angular position  of a line lying in the body. Using the dimensions of the body, relate s to  e.g., s  f ( ) 3. Take the first time derivative of s  f ( ) to get a relationship between v and  4. Take the second time derivative to get a relationship between a and  Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (1 of 3) Given: Link AB is rotating clockwise direction. At the instant   600 the slotted guide rod is moving 2 a  2 m / s to the left with and V  5m / s. Find: The angular velocity and angular acceleration of link AB. Plan: Set the coordinate x to be the distance between A and B. Relate x to the angular position, q. Then take time derivatives of the position equation to find the velocity and acceleration relationships. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (2 of 3) Solution: Relate x, the horizontal distance between A and B, to q. x  0.2cos θ(m) Take time derivatives of the position to find the velocity and acceleration. x  0.2 sin  θ x  0.2cos θ 2  0.2sin  Notice that the velocity and acceleration of the guide rod are directed toward the negative sense of x. When   60 , x   5 m/s and x  2 m/s 2 . Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (3 of 3) Therefore,  5  0.2 sin60     28.9 rad/s and  2  0.2cos 60 2  0.2sin 60  2  0.2cos 6028.9   0.2sin 60 2      470 rad/s 2  470 rad/s 2 Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis: Acceleration (1 of 3) The equation relating the accelerations of two points on the body is determined by differentiating the velocity equation with respect to time. These are absolute accelerations of points A and B. They are measured from a set of fixed x,y axes. The result This term is the acceleration of B with respect to A and includes both tangential and normal components. is a B  a A  (a B/A ) t  (a B/A )n Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis: Acceleration (2 of 3) Graphically: The relative tangential acceleration component ( a B/A ) t is ( a  r B/A ) n and perpendicular to The relative normal acceleration component ( a B/A ) n is (   2 r B/A ) n and the direction is always from B towards A. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Relative Motion Analysis: Acceleration (3 of 3) Since the relative acceleration components can be expressed as ( a B/A ) t    r B/A and ( a B/A ) n    2 r B/A , the relative acceleration equation becomes a B  a A    r B/A   r B/A 2 Note that the last term in the relative acceleration equation is not a cross product. It is the product of a scalar (square of the magnitude of angular velocity,) and the relative position vector, rB/A  2  Square of the magnitude of angular velocity Copyright © 2016 Pearson Education, Inc. All Rights Reserved Application of the Relative Acceleration Equation In applying the relative acceleration equation, the two points used in the analysis (A and B) should generally be selected as points which have a known motion, such as pin connections with other bodies. In this mechanism, point B is known to travel along a circular path, so a B can be expressed in terms of its normal and tangential components. Note that point B on link BC will have the same acceleration as point B on link AB. Point C, connecting link BC and the piston, moves along a straightline path. Hence, a C is directed horizontally. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Procedure for Analysis 1. Establish a fixed coordinate system. 2. Draw the kinematic diagram of the body. 3. Indicate on it a A , a B , ω ,  , and r B/A . If the points A and B move along curved paths, then their accelerations should be indicated in terms of their tangential and normal components, i.e., a A  ( a A ) t  ( a A ) n and a B  ( a B ) t  ( a B ) n . 4. Apply the relative acceleration equation: a B  a A    r B/A   2 r B/A 5. If the solution yields a negative answer for an unknown magnitude, this indicates that the sense of direction of the vector is opposite to that shown on the diagram. Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (1 of 3) Given: Point A on rod AB has an 2 acceleration of 5 m/s and a velocity of 6 m/s at the instant shown. Find: The angular acceleration of the rod and the acceleration at B at this instant. Plan: Follow the problem solving procedure! Solution: First, we need to find the angular velocity of the rod at this instant. Locating the instant center (IC) for rod AB, we can determine    vA/rA/IC  6 / (3)  2 rad/s Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (2 of 3) Since points A and B both move along straight-line paths, aA  5 j m/s 2 aB  aBi m/s 2 Applying the relative acceleration equation a B  a A    r B/A   2 r B/A a B i   5 j  α k  (3 i – 4 j )  2 2 (3 i – 4 j ) a B i   5 j  4 α i  3 α j  (  12 i  16 j )  4 α  12  i  3 α  11  j Copyright © 2016 Pearson Education, Inc. All Rights Reserved Example (3 of 3) So with aB i   (4  – 12) i  (3 a  11 ) j, we can solve for aB and . By comparing the i, j components; aB  4   12 0  3   11 Solving: a B   26 .7 m/s 2  26 .7 m/s 2    3.67 rad/s 2  3.67 rad/s 2 Copyright © 2016 Pearson Education, Inc. All Rights Reserved