Presentation on theme: "ABSOLUTE MOTION ANALYSIS (Section 16.4) Today ’ s Objective: Students will be able to determine the velocity and acceleration of a rigid body undergoing."— Presentation transcript:
ABSOLUTE MOTION ANALYSIS (Section 16.4) Today ’ s Objective: Students will be able to determine the velocity and acceleration of a rigid body undergoing general plane motion using an absolute motion analysis. In-Class Activities: Check homework, if any Reading quiz Applications General Plane Motion Concept quiz Group problem solving Attention quiz
READING QUIZ 1.A body subjected to general plane motion undergoes a/an A)translation. B)rotation. C)simultaneous translation and rotation. D)out of plane movement. 2.In general plane motion, if the rigid body is represented by a slab, the slab rotates A)about an axis perpendicular to the plane. B)about an axis parallel to the plane. C)about an axis lying in the plane. D)None of the above.
APPLICATIONS The position of the piston, x, can be defined as a function of the angular position of the crank, . By differentiating x with respect to time, the velocity of the piston can be related to the angular velocity, , of the crank. 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?
APPLICATIONS (continued) The rolling of a cylinder is an example of general plane motion. During this motion, the cylinder rotates clockwise while it translates to the right. The position of the center, G, is related to the angular position, , by s G = r if the cylinder rolls without slipping. Can you relate the translational velocity of G and the angular velocity of the cylinder?
PROCEDURE FOR ANALYSIS The absolute motion analysis method (also called the parametric method) relates the position of a point, P, on a rigid body undergoing rectilinear motion to the angular position, (parameter), of a line contained in the body. (Often this line is a link in a machine.) Once a relationship in the form of s P = f( ) is established, the velocity and acceleration of point P 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.
Given:Two slider blocks are connected by a rod of length 2 m. Also, v A = 8 m/s and a A = 0. Find:Angular velocity, , and angular acceleration, , of the rod when = 60°. Plan:Choose a fixed reference point and define the position of the slider A in terms of the parameter . Notice from the position vector of A, positive angular position is measured clockwise. EXAMPLE 1
EXAMPLE 1 (continued) Solution: a A = -2 sin – 2 2 cos = 0 = - 2 /tan = rad/s 2 Using ° and v A = 8 m/s and solving for = 8/(-2 sin 60°) = rad/s (The negative sign means the rod rotates counterclockwise as point A goes to the right.) Differentiating v A and solving for , By geometry, s A = 2 cos By differentiating with respect to time, v A = -2 sin A reference sAsA
Given:Crank AB rotates at a constant velocity of = 150 rad/s Find:Velocity of P when = 30° Plan:Define x as a function of and differentiate with respect to time. EXAMPLE 2 v P = -0.2 sin + (0.5)[(0.75) 2 – (0.2sin ) 2 ] -0.5 (-2)(0.2sin )(0.2cos ) v P = -0.2 sin – [0.5(0.2) 2 sin2 ] / (0.75) 2 – (0.2 sin ) 2 At = 30°, = 150 rad/s and v P = ft/s = 18.5 ft/s x P = 0.2 cos +(0.75) 2 – (0.2 sin ) 2 Solution:
CONCEPT QUIZ 1.If the position, s, is given as a function of angular position, , by s = 10 sin 2 , the velocity, v, is A)20 cos 2 B)20 sin 2 C)20 cos 2 D)20 sin 2 2.If s = 10 sin 2 , the acceleration, a, is A)20 sin 2 B) 20 cos 2 - 40 sin 2 C)20 cos 2 D) -40 sin2
GROUP PROBLEM SOLVING Given: The and of the disk and the dimensions as shown. Find:The velocity and acceleration of cylinder B in terms of . Plan:Relate s, the length of cable between A and C, to the angular position, . The velocity of cylinder B is equal to the time rate of change of s.
GROUP PROBLEM SOLVING (continued) Law of cosines: s = (3) 2 + (5) 2 – 2(3)(5) cos v B = (0.5)[34 – 30 cos ] -0.5 (30 sin ) v B = [15 sin ]/ 34 – 30 cos 30cos ) 3/2 - (34 sin ) sin )(30 (-0.5)(15 30cos - 34 sin ) 15 cos (15 2 aBaB cos ) 3/2 30( 2 sin 2 cos ) (34 sin ) 15( 2 cos aBaB Solution:
ATTENTION QUIZ 2.If v A =10 m/s, determine the angular acceleration, when = 30 . A) 0 rad/s 2 B) rad/s 2 C) -112 rad/s 2 D)-173 rad/s 2 v A =10 m/s 1.The sliders shown below are confined to move in the horizontal and vertical slots. If v A =10 m/s, determine the connecting bar ’ s angular velocity when = 30 . A)10 rad/sB)10 rad/s C)8.7 rad/sD)8.7 rad/s