ENGR DYNAMICS. Rigid-body Lecture 4. Relative motion analysis: acceleration. Acknowledgements

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1 ENGR DYNAMICS Rigid-body Lecture 4 Relative motion analysis: acceleration Acknowledgements These lecture slides were provided by, and are the copyright of, Prentice Hall (*) as part of the online instructor resources for the textbook: Engineering Mechanics: Dynamics, by R.C. Hibbler, 11th Edition, PEARSON/Prentice Hall (*) except for some modifications performed by Dr. ( ) except for some modifications performed by Dr. Gilberto E. Urroz for this course 1

RELATIVE MOTION ANALYSIS: ACCELERATION Today s Objectives: Students will be able to: 1. Resolve the acceleration of a point on a body into components of translation and rotation. 2.

2 RELATIVE MOTION ANALYSIS: ACCELERATION Today s Objectives: Students will be able to: 1. Resolve the acceleration of a point on a body into components of translation and rotation. 2. Determine the acceleration of a point on a body by using a relative acceleration analysis. In-Class Activities: Check Homework Reading Quiz Applications Translation and Rotation Components of Acceleration Relative Acceleration Analysis Roll-Without-Slip Motion Concept Quiz Group Problem Solving Attention Quiz READING QUIZ 1. If two bodies contact one another without slipping, and the points in contact move along different paths, the tangential components of acceleration will be and the normal components of acceleration will be. A) the same, the same B) the same, different C) different, the same D) different, different 2. When considering a point on a rigid body in general plane motion, A) It s total acceleration consists of both absolute acceleration and relative acceleration components. B) It s total acceleration consists of only absolute acceleration components. C) It s relative acceleration component is always normal to the path. D) None of the above. 2

APPLICATIONS In the mechanism for a window, link AC rotates about a fixed axis through C, while point B

The components of acceleration of these points can be inferred since their motions are known.

How can we determine the accelerations of the links in the mechanism?

3 APPLICATIONS In the mechanism for a window, link AC rotates about a fixed axis through C, while point B slides in a straight track. The components of acceleration of these points can be inferred since their motions are known. To prevent damage to the window, the accelerations of the links must be limited. How can we determine the accelerations of the links in the mechanism? APPLICATIONS The forces delivered to the crankshaft, and the angular acceleration of the crankshaft, depend on the speed and acceleration of the piston in an automotive engine. How can we relate the accelerations of the piston, connection rod, and crankshaft in this engine? 3

RELATIVE MOTION ANALYSIS: ACCELERATION (Section 16-7) The equation relating the accelerations of two points on the body is

dv B dt dv A dt dv B / A dt These are absolute accelerations of points A and B. They are measured from a set of fixed x,y axes.

The result is a B = a A + (a B/A ) t + (a B/A ) n RELATIVE MOTION ANALYSIS: ACCELERATION Graphically: a B = a A + (a B/A ) t +(a

4 RELATIVE MOTION ANALYSIS: ACCELERATION (Section 16-7) The equation relating the accelerations of two points on the body is determined by differentiating the velocity equation with respect to time. dv B dt dv A dt dv B / A dt These are absolute accelerations of points A and B. They are measured from a set of fixed x,y axes. This term is the acceleration of B with respect to A. It will develop tangential and normal components. The result is a B = a A + (a B/A ) t + (a B/A ) n RELATIVE MOTION ANALYSIS: ACCELERATION Graphically: a B = a A + (a B/A ) t +(a B/A ) n = + The relative tangential acceleration component (a B/A ) t is ( x r B/A ) and perpendicular to r B/A. The relative normal acceleration component (a B/A ) n is (- 2 r B/A ) and the direction is always from B towards A. 4

RELATIVE MOTION ANALYSIS: ACCELERATION Since the relative acceleration components can be expressed as (a = =- 2 B/A ) t r B/A and (a B/A ) n r B/A the relative acceleration equation becomes a B = a A

5 RELATIVE MOTION ANALYSIS: ACCELERATION Since the relative acceleration components can be expressed as (a = =- 2 B/A ) t r B/A and (a B/A ) n r B/A the relative acceleration equation becomes a B = a A + r B/A - 2 r B/A 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, 2 ) and the relative position vector, r B/A. APPLICATION OF 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 ti 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 straight-line path. Hence, a C is directed horizontally. 5

6 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,,, 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, it indicates the sense of direction of the vector is opposite to that shown on the diagram. EXAMPLE Plan: Given:Point A on rod AB has an acceleration of 3 m/s 2 and a velocity of 2 m/s at the instant tthe rod becomes horizontal. Find: The angular acceleration of the rod at this instant. Follow the problem solving procedure! Slti Solution: First, we need to find the angular velocity of the rod at this instant. Locating the instant center (IC) for rod AB (which lies above the midpoint of the rod), we can determine : A = v A /r A/IC = v A /(5/cos 45) = rad/s 6

7 EXAMPLE Since points A and B both move along straight-line paths, a A = 3 (cos 45 i - sin 45 j) m/s a B = a B (cos 45 i + sin 45 j) m/s Applying the relative acceleration equation a B = a A + x r B/A 2 r B/A (a B cos 45 i + a B sin 45 j = (3 cos 45 i 3 sin 45 j) + ( k x 10 i) (0.283) 2 (10i) EXAMPLE By comparing the i, j components; a B cos 45 = 3 cos 45 (0.283) 2 ((10) a B sin 45 = -3 sin 45 + (10) Solving: a B = 1.87 m/s 2 = rad/s 2 7

BODIES IN CONTACT Consider two bodies in contact with one another without slipping, where the points in contact move

, (a A ) t = (a A ) t (which implies B r B = C r C ). The normal components of acceleration will not be the same.

motion without slip; e.g., a ball or disk rolling along a flat surface without slipping.

A th li d ll i tg ( t ) l t i htli As the cylinder rolls, point G (center) moves along a straight line, while point A,

8 BODIES IN CONTACT Consider two bodies in contact with one another without slipping, where the points in contact move along different paths. In this case, the tangential components of acceleration will be the same, i. e., (a A ) t = (a A ) t (which implies B r B = C r C ). The normal components of acceleration will not be the same. (a A ) n (a A ) n so a A a A ROLLING MOTION Another common type of problem encountered in dynamics involves rolling motion without slip; e.g., a ball or disk rolling along a flat surface without slipping. This problem can be analyzed using relative velocity and acceleration equations. A th li d ll i tg ( t ) l t i htli As the cylinder rolls, point G (center) moves along a straight line, while point A, on the rim of the cylinder, moves along a curved path called a cycloid. If and are known, the relative velocity and acceleration equations can be applied to these points, at the instant A is in contact with the ground. 8

9 Velocity: Acceleration: ROLLING MOTION Since no slip occurs, v A = 0 when A is in contact with ground. From the kinematic diagram: v G = v A + x r G/A v G i = 0 + (- k) x (r j) v G = r or v G = r i Since G moves along a straight-line path, a G is horizontal. Just before A touches ground, its velocity is directed downward, and just after contact, its velocity is directed upward. Thus, point A accelerates upward as it leaves the ground. a G = a A + x r G/A 2 r G/A => a G i = a A j + (- k) x (r j) 2 (r j) Evaluating and equating i and j components: a G = r and a A = 2 r or a G = r i and a A = 2 r j EXAMPLE II Given:The ball rolls without slipping. Find: The accelerations of points A and B at this instant. Plan: Follow the solution procedure. Solution: Since the ball is rolling without slip, a O is directed to the left with a magnitude of: a O = r = (4 rad/s 2 )(0.5 ft)=2 ft/s 2 9

10 EXAMPLE II Now, apply the relative acceleration equation between points O and B. a B = a O + x r B/O 2 r B/O Now do the same for point A. a B = -2i + (4k) x (0.5i) (6) 2 (0.5i) = (-20i + 2j) ft/s 2 a A = a O + x r A/O 2 r A/O a A = -2i + (4k) x (0.5i) (6) 2 (0.5j) = (-4i 18j) ft/s 2 CONCEPT QUIZ 1. If a ball rolls without slipping, select the tangential and normal components of the relative acceleration of point A with respect to G. A) r i + 2 r j B) - r i + 2 r j C) 2 r i - r j D) Zero. y x B 2. What are the tangential and normal components of the relative acceleration of point B with respect to G. A) - 2 r i - r j B) - r i + 2 r j C) 2 r i - r j D) Zero. 10

GROUP PROBLEM SOLVING Given: The disk is rotating with = 3 rad/s, = 8 rad/s 2 at this instant. Find: The acceleration at point B, and the angular velocity and acceleration of link AB.

11 GROUP PROBLEM SOLVING Given: The disk is rotating with = 3 rad/s, = 8 rad/s 2 at this instant. Find: The acceleration at point B, and the angular velocity and acceleration of link AB. Plan: Follow the solution procedure. Solution: At the instant shown, points A and B are both moving horizontally. Therefore, link AB is translating, meaning AB = 0. GROUP PROBLEM SOLVING (a A ) t Draw the kinematic diagram and then AB apply the relative-acceleration (a A ) n equation: a B = a A + AB x r B/A 2 ABr B/A Where a An = 0.2*3 2 = 1.8 m/s 2 a At = 0.2*8 = 1.6 m/s 2 r B/A = 0.4 cos 30i 0.4 sin 30j AB = AB k, w AB = 0 a B i = (1.6i 1.8j) + AB k x (0.4 cos 30i 0.4 sin 30j) = ( sin 30 AB )i + ( cos 30 AB )j 11

GROUP PROBLEM SOLVING a At By comparing i, j components: a An AB a B = 16 1.6 + 04 0.4 sin 30 AB 0 = -1.8 + 0.4 cos 30 AB Solving: a B = 2.64 m/s 2 AB = 5.20 rad/s 2 ATTENTION QUIZ 1.

12 GROUP PROBLEM SOLVING a At By comparing i, j components: a An AB a B = sin 30 AB 0 = cos 30 AB Solving: a B = 2.64 m/s 2 AB = 5.20 rad/s 2 ATTENTION QUIZ 1. Two bodies contact one another without slipping. If the tangential component of the acceleration of point A on gear B is 100 ft/sec 2, dt determine the tangential component of the acceleration of point A on gear C. A) 50 ft/sec 2 B) 100 ft/sec 2 C) 200 ft/sec 2 D) None of above. 2 If h i l f h l i f i A 2. If the tangential component of the acceleration of point A on gear B is 100 ft/sec 2, determine the angular acceleration of gear B. A) 50 rad/sec 2 B) 100 rad/sec 2 C) 200 rad/sec 2 D) None of above. 2 ft 1 ft 12

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CEE 271: Applied Mechanics II, Dynamics Lecture 23: Ch.16, Sec.7

1 / 26 CEE 271: Applied Mechanics II, Dynamics Lecture 23: Ch.16, Sec.7 Prof. Albert S. Kim Civil and Environmental Engineering, University of Hawaii at Manoa Tuesday, Nov. 8, 2012 2 / 26 RELATIVE MOTION