Dynamics Lecture5 Dependent Motion of Two Bodies and Relative Motion Using Translating Axes

©2007 Pearson Education South Asia Pte LtdAbsolute Dependent Motion Analysis of Two Particles • Motion of one particle will depend on the corresponding motion of another particle • Dependency occurs when particles are interconnected by the inextensible cords which are wrapped around pulleys • For example, the movement of block A downward along the inclined plane will cause a corresponding movement of block B up the other incline • Specify the locations of the blocks using position coordinate s A and s B ©2007 Pearson Education South Asia Pte Ltd • Note each of the coordinate axes is (1) referenced from a fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of block A and block B and (3) has a positive sense from C to A and D to B • If total cord length is lT, the position coordinate are related by the equation T B CD A l s l s = + + Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd • Here l CD is the length passing over arc CD • Taking time derivative of this expression, realizing that l CD and l T remain constant, while s A and s B measure the lengths of the changing segments of the cord A B B A v v dt ds dt ds ÷ = = + 0 or • The negative sign indicates that when block A has a velocity downward in the direction of position s A , it causes a corresponding upward velocity of block B; B moving in the negative s B direction Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd • Time differentiation of the velocities yields the relation between accelerations a B = - a A • For example involving dependent motion of two blocks • Position of block A is specified by s A , and the position of the end of the cord which block B is suspended is defined by s B Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd • Choose coordinate axes which are (1) referenced from fixed points and datums, (2) measured in the direction of motion of each block, (3) positive to the right (s A ) and positive downward (s B ) • During the motion, the red colored segments of the cord remain constant • If l represents the total length of the cord minus these segments, then the position coordinates can be related by l s h s A B = + + 2 Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd Since l and h are constant during the motion, the two time derivatives yields A B A B a a v v ÷ = ÷ = 2 2 • When B moves downward (+s B ), A moves to left (-s A ) with two times the motion • This example can also be worked by defining the position of block B from the center of the bottom pulley ( a fixed point) Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd A B A B A B a a v v l s h s h = = = + + ÷ 2 2 ) ( 2 - Time differentiation yields Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd PROCEDURE FOR ANALYSIS Position-Coordinate Equation • Establish position coordinates which have their origin located at a fixed point or datum • The coordinates are directed along the path of motion and extend to a point having the same motion as each of the particles • It is not necessary that the origin be the same for each of the coordinates; however, it is important that each coordinate axis selected be directed along the path of motion of the particle Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd • Using geometry or trigonometry, relate the coordinates to the total length of the cord, l T , or to that portion of cord, l, which excludes the segments that do not change length as the particles move – such as arc segments wrapped over pulleys • For problem involving a system of two or more cords wrapped over pulleys, then the position of a point on one cord must be related to the position of a point on another cord using the above procedure • Separate equations must be written for a fixed length of each cord of the system. Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd Time Derivatives • Two successive time derivatives of the position- coordinates equations yield the required velocity and acceleration equations which relate motions of the particles • The signs of the terms in these equations will be consistent with those that specify the positive and negative sense of the position coordinates Absolute Dependent Motion Analysis of Two Particles ©2007 Pearson Education South Asia Pte Ltd EXAMPLE 12.21 Determine the speed of block A if block B has an upward speed of 2 m/s. ©2007 Pearson Education South Asia Pte Ltd Position Coordinate System. There is one cord in this system having segments which are changing length. Position coordinates s A and s B will be used since each is measured from a fixed point (C or D) and extends along each block’s path of motion. In particular, s B is directed to point E since motion of B and E is the same. The red colored segments of the cords remain at a constant length and do not have to be considered as the block move. EXAMPLE 12.21 ©2007 Pearson Education South Asia Pte Ltd The remaining length of the cord, l, is also considered and is related to the changing position coordinates s A and s B by the equation l s s B A = + 3 Time Derivative. Taking the time derivative yields 0 3 = + B A v v so that when v B = -2m/s (upward) v A = 6m/s ↓ EXAMPLE 12.21 ©2007 Pearson Education South Asia Pte Ltd Determine the speed of block A if block B has an upward speed of 2m/s. EXAMPLE 12.22 ©2007 Pearson Education South Asia Pte Ltd Position Coordinate Equation. Positions of A and B are defined using coordinates s A and s B . Since the system has two cords which change length, it is necessary to use a third coordinate s C in order to relate s A to s B . Length of the cords can be expressed in terms of s A and s C , and the length of the other cord can be expressed in terms of s B and s C . The red colored segments are not considered in this analysis. EXAMPLE 12.22 ©2007 Pearson Education South Asia Pte Ltd For the remaining cord length, 2 1 ) ( 2 l s s s l s s C B B C A = ÷ + = + Eliminating sC yields, 1 2 2 4 l l s s B A + = + Time Derivative. The time derivative gives 0 4 = + B A v v so that v B = -2m/s (upward) + = + = s m s m v B / 8 / 8 EXAMPLE 12.22 ©2007 Pearson Education South Asia Pte Ltd Determine the speed with which block B rises if the end of the cord at A is pulled down with a speed of 2m/s. EXAMPLE 12.23 ©2007 Pearson Education South Asia Pte Ltd Position-Coordinate Equation. The position of A is defined by s A , and the position of block B is specified by s B since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fixed pin at pulley D. Since the system consists of two cords, the coordinates s A and s B cannot be related directly. By establishing a third position coordinate, s C , and the length of the other cord in terms of s A , s B and s C . EXAMPLE 12.23 ©2007 Pearson Education South Asia Pte Ltd Excluding the red colored segments of the cords, the remaining constant cord lengths l 1 and l 2 (along the hook and link dimensions) can be expressed as 1 2 2 4 l l s s B C + = + ( ) ( ) 2 1 l s s s s s l s s B C B C A B C = + ÷ + ÷ = + Eliminating s C yields EXAMPLE 12.23 ©2007 Pearson Education South Asia Pte Ltd Time Derivative. The time derivative give | = ÷ = = + s m s m v v v B B A / 5 . 0 / 5 . 0 0 4 when v A = 2m/s (downward) EXAMPLE 12.23 ©2007 Pearson Education South Asia Pte Ltd A man at A is hoisting a safe S by walking to the right with a constant velocity v A = 0.5m/s. Determine the velocity and acceleration of the safe when it reaches the elevation at E. The rope is 30m long and passes over a small pulley at D. EXAMPLE 12.24 ©2007 Pearson Education South Asia Pte Ltd Position Coordinate System. Rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the position of S and A, are specified by means of the x and y coordinates measured from a fixed point and directed along the paths of motion of the ends of the rope. The x and y coordinates may be related since the rope has a fixed length l = 30m, which at all times is equal to the length of the segment DA plus CD. EXAMPLE 12.24 View Free Body Diagram ©2007 Pearson Education South Asia Pte Ltd Using Pythagorean Theorem and the equation for lCD to get the relationship between y and x: ( ) ( ) ) 2 . ( 15 ) 1 . ( 15 2 2 eq y l eq x l CD DA ÷ = + = ( ) ( ) 15 225 15 15 30 ) 3 . ( 2 2 2 ÷ + = ÷ + + = + = x y y x eq l l l CD DA (eq.4) EXAMPLE 12.24 ©2007 Pearson Education South Asia Pte Ltd Time Derivative. Taking time derivative, using the chain rule where, v S = dy/dt and v A = dx/dt A S v x x dt dx x x dt dy v 2 2 225 225 2 2 1 + = ( ¸ ( ¸ + = = At y = 10 m, x = 20 m, v A = 0.5 m/s, v S = 400mm/s ↑ (2) EXAMPLE 12.24 ©2007 Pearson Education South Asia Pte Ltd The acceleration is determined by taking the time derivative of eqn (2), ( ) 2 / 3 2 2 2 2 2 / 3 2 2 2 225 225 225 1 225 1 ) 225 ( ) / ( x v dt dv x x v dt dx x xv x dt dx x dt y d a A A A A S + = ( ¸ ( ¸ + + | . | \ | ( ¸ ( ¸ + + ( ¸ ( ¸ + ÷ = = At x = 20 m, with v A = 0.5 m/s, | = 2 / 6 . 3 s mm a S EXAMPLE 12.24 ©2007 Pearson Education South Asia Pte Ltd Relative Motion Analysis of Two Particles Using Translating Axes • There are many cases where the path of the motion for a particle is complicated, so that it may be feasible to analyze the motions in parts by using two or more frames of reference • For example, motion of an particle located at the tip of an airplane propeller while the plane is in flight, is more easily described if one observes first the motion of the airplane from a fixed reference and then superimposes (vectorially) the circular motion of the particle measured from a reference attached to the airplane ©2007 Pearson Education South Asia Pte Ltd Position. • Consider particle A and B, which moves along the arbitrary paths aa and bb, respectively • The absolute position of each particle r A and r B , is measured from the common origin O of the fixed x, y, z reference frame Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd • Origin of the second frame of reference x’, y’ and z’ is attached to and moves with particle A • Axes of this frame only permitted to translate relative to fixed frame • Relative position of “B with respect to A” is designated by a relative-position vector r B/A • Using vector addition A B A B r r r /    + = Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd Velocity. • By time derivatives, • Here refer to absolute velocities, since they are observed from the fixed frame • Relative velocity is observed from the translating frame A B A B v v v /    + = dt r d v and dt r d v A A B B / /     = = dt r d v A B A B / / /   = Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd • Since the x’, y’ and z’ axes translate, the components of r B/A will not change direction and therefore time derivative o this vector components will only have to account for the change in the vector magnitude • Velocity of B is equal to the velocity of A plus (vectorially) the relative velocity of “B relative to A” as measured by the translating observer fixed in the x’, y’ and z’ reference Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd Acceleration. • The time derivative yields a similar relationship between the absolute and relative accelerations of the particles A and B • Here a B/A is the acceleration of B as seen by the observer located at A and translating with the x’, y’ and z’ reference frame A B A B a a a /    + = Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd PROCEDURE FOR ANALYSIS • When applying the relative position equations, r B = r A + r B/A , it is necessary to specify the location of the fixed x, y , z and translating x’, y’ and z’ • Usually, the origin A of the translating axes is located at a point having a known position r A • A graphical representation of the vector addition can be shown, and both the known and unknown quantities labeled on this sketch Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd • Since vector addition forms a triangle, there can be at most two unknowns, represented by the magnitudes and/or directions of the vector quantities • These unknown can be solved for either graphically, using trigonometry, or resolving each of the three vectors r A , r B and r B/A into rectangular or Cartesian components, thereby generating a set of scalar equations Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd • The relative motion equations v B = v A + v B/A and a B = a A + a B/A are applied in the same manner as explained above, except in this case, origin O of the fixed axes x, y, z axes does not have to be specified Relative Motion Analysis of Two Particles Using Translating Axes ©2007 Pearson Education South Asia Pte Ltd A train, traveling at a constant speed of 90km/h, crosses over a road. If automobile A is traveling t 67.5km/h along the road, determine the magnitude and direction of relative velocity of the train with respect to the automobile EXAMPLE 12.25 ©2007 Pearson Education South Asia Pte Ltd Vector Analysis. The relative velocity is measured from the translating x’, y’ axes attached to the automobile. Since v T and v A are known in both magnitude and direction, the unknowns become the x and y components of v T/A . Using the x, y axes and a Cartesian vector analysis, h km j i v v j i i v v v A T A T A T A T / ) ~ 7 . 47 ~ 3 . 42 { ) ~ 45 sin 5 . 67 ~ 45 cos 5 . 67 ( ~ 90 / / / ÷ = + + = + =        EXAMPLE 12.25 ©2007 Pearson Education South Asia Pte Ltd The magnitude of v T/A is h km v A T / 8 . 63 ) 7 . 47 3 . 42 ( 2 2 / = + = The direction of v T/A defined from the x axis is ( ) ( )  40 . 48 3 . 42 7 . 47 tan / / = = = u u x A T y A T v v EXAMPLE 12.25 ©2007 Pearson Education South Asia Pte Ltd Plane A is flying along a straight-line path, while plane B is flying along a circular path having a radius of curvature of ρ B = 400 km. Determine the velocity and acceleration of B as measured by the pilot of A. EXAMPLE 12.26 ©2007 Pearson Education South Asia Pte Ltd Velocity. The x, y axes are located at an arbitrary fixed point. Since the motion relative to plane A is to be determined, the translating frame of reference x’. y’ is attached to it. Applying the relative-velocity equation in scalar form since the velocity vectors of both plane are parallel at the instant shown, + = ÷ = + = + = h km h km v v v v v A B A B A B A B / 100 / 100 700 600 / / / ) ( | + EXAMPLE 12.26 ©2007 Pearson Education South Asia Pte Ltd Acceleration. Plane B has both tangential and normal components of acceleration, since it is flying along a curved path. Magnitude of normal acceleration, ( ) 2 2 / 900 h km v a B n B = = µ Applying the relative-acceleration equation, { } 2 / / / / ~ 150 ~ 900 ~ 50 ~ 100 ~ 900 h km j i a a j j i a a a A B A B A B A B ÷ = + = ÷ + = EXAMPLE 12.26 ©2007 Pearson Education South Asia Pte Ltd From the figure shown, the magnitude and direction of A B a /  46 . 9 900 150 tan / 912 1 2 / = = = ÷ u h km a A B EXAMPLE 12.26 ©2007 Pearson Education South Asia Pte Ltd At the instant, car A and B are traveling with the speed of 18 m/s and 12 m/s respectively. Also at this instant, A has a decrease in speed of 2 m/s 2 , and B has an increase in speed of 3 m/s 2 . Determine the velocity and acceleration of B with respect to A. EXAMPLE 12.27 ©2007 Pearson Education South Asia Pte Ltd Velocity. The fixed x, y axes are established at a point on the ground and the translating x’, y’ axes are attached to car A. Using Cartesian vector analysis, ( ) { } s m v s m j i v v j i j v v v A B A B A B A B A B / 69 . 9 588 . 3 9 / ~ 588 . 3 ~ 9 ~ 60 sin 18 ~ 60 cos 18 ~ 12 2 2 / / / / = + = + = + ÷ ÷ = ÷ + =   Thus, EXAMPLE 12.27 View Free Body Diagram ©2007 Pearson Education South Asia Pte Ltd Its direction is ( ) ( )  7 . 21 9 588 . 3 tan / / = = = u u x A B y A B v v Acceleration. The magnitude of the normal component is ( ) 2 2 / 440 . 1 s m v a B n B = = µ EXAMPLE 12.27 ©2007 Pearson Education South Asia Pte Ltd Applying the equation for relative acceleration yields ( ) ( ) { } 2 / / / / ~ 732 . 4 ~ 440 . 2 ~ 60 sin 2 ~ 60 cos 2 ~ 3 ~ 440 . 1 s m j i a a j i j i a a a A B A B A B A B ÷ ÷ = + + = ÷ ÷ + =   Magnitude and direction is  7 . 62 / 32 . 5 2 / = = | s m a A B EXAMPLE 12.27