Understanding Relative Velocity of Bodies in Motion

The content explains concepts related to relative velocity of moving bodies, including diagrams illustrating velocity relationships, application of laws of parallelogram and triangle, analysis of motion in rigid links, and calculation of rubbing velocity at pin joints in mechanisms. It covers scenarios of bodies moving in straight lines, inclined directions, and on rotating links, providing a comprehensive insight into how relative velocities are determined and represented. The discussions delve into both theoretical explanations and practical applications of relative motion in various mechanical systems.

• Relative Velocity
• Motion Diagrams
• Rigid Links
• Rubbing Velocity
• Mechanisms

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1. First Semester chapter one Velocity and acceleration diagram Dr. Adil Bed Nayeeif 2019-2020

2. Relative velocity of two bodies moving in straight line Consider two bodies A and B moving along parallel lines in the same direction with absolute velocities VA and VB such that VA > VB , as shown in Figure (1). The relative velocity of A with respect to B, VAB = Vector difference of VA and VB = VA VB From Figure (2), the relative velocity of A with respect to B (i.e. VAB) may be written in the vector form as follows: ba = oa ob Figure (1) Figure (2)

3. consider the body B moving in an inclined direction as shown in Figure (3). The relative velocity of A with respect to B may be obtained by the law of parallelogram of velocities or triangle law of velocities. Take any fixed point o and draw vector oa to represent VA in magnitude and direction to some suitable scale. Similarly, draw vector ob to represent VB in magnitude and direction to the same scale. Then vector ba represents the relative velocity of A with respect to B as shown in Figure(4). In the similar way as discussed above, the relative velocity of A with respect to B, vAB = Vector difference of vA and vB = vA vB or ba = oa ob Figure (3) Figure (4)

4. Motion of a link Consider two points A and B on a rigid link AB, as shown in Figure(5). Let one of the extremities (B) of the link move relative to A, in a clockwise direction. Since the distance from A to B remains the same, therefore there can be no relative motion between A and B, along the line AB. It is thus obvious, that the relative motion of B with respect to A must be perpendicular to AB. The relative velocity of B with respect to A (i.e. VAB) is represented by the vector ab and is perpendicular to the line AB as shown in Figure(6). = Angular velocity of the link AB about A. We know that the velocity of the point B with respect to A, VBA= ab = . AB Similarly, the velocity of any point C on AB with respect to A, VCA= ac = . AC Figure (5) Figure (6)

5. Velocity diagram for a block sliding on a rotating link Let ( ) be the angular of the link about the fixed point (o), and VA be velocity of the block, assumed known in magnitude and direction. If ? is the point on the link coincident with the block, the velocity of ? relative to o shown in Figure (7). Figure (7)

6. Rubbing velocity at a pin joint The links in a mechanism are mostly connected by means of pin joints. The rubbing velocity is defined as the algebraic sum between the angular velocities of the two links which are connected by pin joints, multiplied by the radius of the pin. Consider two links OA and OB connected by a pin joint at O as shown in Figure (8). Let 1 = Angular velocity of the link OA or the angular velocity of the point A with respect to O. 2 = Angular velocity of the link OB or the angular velocity of the point B with respect to O, and r = Radius of the pin. According to the definition, Rubbing velocity at the pin joint O = ( 1 2) r, if the links move in the same direction = ( 1 + 2) r, if the links move in the opposite direction Figure (8)

7. Analytical determination of piston velocity sin sin L n where n L = + + ( ) ( cos cos ) X r L r L = r L sin r = = = sin sin r 2 sin sin 2 = cos 1 1 2 n n 2 1 n sin ce is small sin 2 = + 1 ( r cos ) X L 2 n 2 sin 2 = + 1 ( r cos ) X 2 n 2 sin 2 n dx d = = + (sin ) V r P 2 dt dt sin 2 = + (sin ) V r P n cos 2 d x d 2 = = + (cos ) a r P dt n dt 2 cos 2 = + (cos ) a r 2 P n

8. Coriolis component of acceleration This tangential component of acceleration of the slider B with respect to the coincident point C on the link is known as Coriolis component of acceleration and is always perpendicular to the link. Coriolis component of the acceleration of B with respect of C, aC = at = 2 .v = Angular velocity of the link OA, and v = Velocity of slider B with respect to coincident point C. The anticlockwise direction for and the radially outward direction for v are taken as positive. It may be noted that the direction of Coriolis component of acceleration changes sign, if either or v is reversed in direction. But the direction of Coriolis component of acceleration will not be changed in sign if both and v are reversed in direction. It is concluded that the direction of Coriolis component of acceleration is obtained by rotating v, at 90 , about its origin in the same direction as that of .

9. Examples EX(1) An engine mechanism is shown in Figure. The crank CB = 100 mm and the connecting rod BA = 300 mm with centre of gravity G,100 mm from B. In the position shown, the crankshaft has a speed of 75 rad/s and an angular acceleration of 1200 rad/s2. Find:1. velocity of G and angular velocity of AB, and 2. acceleration of G and angular acceleration of AB. Solution. Given : BC = 75 rad/s ; BC = 1200 rad/s2, CB = 100 mm = 0.1 m; BA = 300 mm= 0.3 m We know that velocity of B with respect to C or velocity of B, vBC= vB= BC. CB = 75 . 0.1 = 7.5 m/s Since the angular acceleration of the crankshaft, BC = 1200 rad/s2, therefore tangential component of the acceleration of B with respect to C, atBC = BC BC = 1200 0.1=120 m/s draw the space diagram, to some suitable scale, as shown in Figure (9). Figure (9)

10. Velocity diagram Draw vector cb perpendicular to CB, to some suitable scale, to represent the velocity of B with respect to C or velocity of B (i.e. vBC or vB), such that vector cb = vBC = vB = 7.5 m/s . 1. From point b, draw vector ba perpendicular to BA to represent the velocity of A with respect to B i.e. vAB, and from point c, draw vector ca parallel to the path of motion of A (which is along AC) to represent the velocity of A i.e. vA. The vectors ba and ca intersect at a. 2. Since the point G lies on AB, therefore divide vector ab at g in the same ratio as G divides AB in the space diagram. In other words, ag /ab = AG/ AB. 3. The vector cg represents the velocity of G. By measurement, we find that velocity of G, vG = vector cg = 6.8 m/s Ans From velocity diagram, we find that velocity of A with respect to B, vAB= vector ba = 4 m/s We know that angular velocity of AB, AB= VAB/AB= 4/0.3= 13.3 rad/ sec (Clockwise)Ans.

11. Acceleration diagram We know that radial component of the acceleration of B with respect to C, ( ) sec / 5 . 562 1 . 0 CB ( ) 2 sec / 3 . 53 3 . 0 BA 2 5 . 7 v 2 BC = = = a m 2 r BC and radial component of the acceleration of A with respect to B, 2 2 4 m a AB = = = v r AB The acceleration diagram, as shown in Figure(10) , is drawn as discussed below: Draw vector c' b'' parallel to CB, to some suitable scale, to (c) Acceleration diagram represent the radial component of the acceleration of B with respect to C, i.e. BC, ar such that vector c b BC 562.5 m/s c b = 1. a r BC = 562.5 m/sec2. From point b'', draw vector b'' b' perpendicular to vector c' b'' or CB to represent the tangential component of the acceleration of B with respect to C i.e. BC at , such that vector b b = = 120 m/s2 . a 2. t BC Join c' b'. The vector c' b' represents the total acceleration of B with respect to C i.e. aBC . 3. a r AB From point b', draw vector b' x parallel to BA to represent radial component of the acceleration of A with respect to B i.e. that vector b x = = 53.3 m/s2. AB a such 4. r From point x, draw vector xa' perpendicular to vector b'x or BA to represent tangential. component of the acceleration of A with respect to B i.e. , whose magnitude is not yet known. AB a 5. t Now draw vector c' a' parallel to the path of motion of A (which is along AC) to represent the acceleration of A i.e. aA. The vectors xa' and c'a' intersect at a'. Join b' a'. The vector b' a' represents the acceleration of A with respect to B i.e. aAB. 6. In order to find the acceleration of G, divide vector a' b' in g' in the same ratio as G divides BA in Figure. Join c' g'. The vector c' g' represents the acceleration of G. 7.

12. Acceleration diagram By measurement, we find that acceleration of G, aA = vector c' g' = 414 m/s2Ans. From acceleration diagram, we find that tangential component of the acceleration of A with respect to B, = vector xa = 546 m/s2 AB a t 546 a t AB = = = 1820 / rad s 2 Angular acceleration of AB, (Clockwise) Ans. BA AB 3 . 0 Figure (10)

13. Ex(2) The dimensions and configuration of the four bar mechanism, shown in Figure (11) are as follows : P1A = 300 mm; P2B = 360 mm; AB 360mm, and P1P2 = 600 mm. The angle AP1 P2 = 60 . The crank P1A has an angular velocity of 10 rad/s and an angular acceleration of 30 rad/s2, both clockwise. Determine the angular velocities and angular accelerations of P2B, and AB and the velocity and acceleration of the joint B. SolutionGiven : AP1 = 10 rad/s ; AP1 = 30 rad/s2; P1A = 300 mm = 0.3 m ; P2B = AB =360 mm = 0.36 m We know that the velocity of A with respect to P1 or velocity of A, VAp1= VA= AP1 P1A= 10 0.3 = 3 m/s Velocity of B and angular velocities of P 2B and AB Draw the space diagram, to some suitable scale, as shown in Figure(a). The velocity diagram, as shown in Figure(b), is drawn as discussed below: Figure (11) Since P1 and P2 are fixed points, therefore these points lie at one place in velocity diagram. 1. Draw vector p1 a perpendicular to P1A, to some suitable scale, to represent the velocity of A with respect to P1 or velocity of A i.e. vAP1 or vA, such that vector p1a = vA1P= vA = 3 m/s. From point a, draw vector ab perpendicular to AB to represent velocity of B with respect to A (i.e. vBA ) and from point p2 draw vector p2b perpendicular to P2B to represent the velocity of B with respect to P2 or velocity of i.e. vBP2 or vB. The vectors ab and p2b intersect at b.By measurement, we find that. 2. vBP2 = vB = vector p2b = 2.2 m/s Ans. ,and vBA = vector ab = 2.05 m/s 3. v 2 . 2 = = = 1 . 6 / sec rad We know that angular velocity of P2B, (Anticlockwise) Ans. 2 B P 05 . 2 AB 2 p B . 0 36 P B 2 and angular velocity of AB, (Anticlockwise) Ans . 0 vBA = = 7 . 5 = / sec rad AB 36

14. Acceleration of B and angular acceleration of P2B and AB = = 3 . 0 = 30 9 / P A rad s We know that tangential component of the acceleration of A with respect to P1, 2 t AP 1 1 1 AP Radial component of the acceleration of A with respect to P1, 2 1 1 10 A P A P v = = = 3 . 0 = 30 / a m s 2 AP 2 2 r AP AP ( . 2 ) 1 1 2 05 v 2 BA = = = 1 11 67 . / a m s 2 r BA 1 Radial component of the acceleration of B with respect to A. . 0 36 AB and radial component of the acceleration of B with respect to P2, ( ) 2 2 44 . 13 36 . 0 B P 2 2 . 2 v 2 BP = = = / a m s 2 r BP 2 The acceleration diagram, as shown in Figure(c), is drawn as follows: Since P1and P2 are fixed points, therefore these points will lie at one place, in the acceleration diagram. Draw vector p1' x parallel to P1A, to some suitable scale, to represent the radial component of the acceleration of A with respect to P1,such that vector p x = =30 m/s2. AP1 1. r From point x, draw vector xa'perpendicular to P1A to represent the tangential component of the acceleration of A with respect to P1, such that vector xa = = 9 m/s2. AP1 2. t Join p1' a'. The vector p1' a' represents the acceleration of A. By measurement, we find that the acceleration of A, aA = aAP1 = 31.6 m/s2 . 3. From point a', draw vector a' y parallel to AB to represent the radial component of the acceleration of B with respect to A, such that vector a y = =11.67 m/s2. AB a 4. r

15. 5.From point y, draw vector yb' perpendicular to AB to represent the tangential component of the acceleration of B with respect to A (i.e. ) whose magnitude is yet unknown. BA a t 6. Now from point p2 , draw vector p2 z parallel to P2B to represent the radial component of the acceleration B with respect to P2, such that vector p z = = 13.44 m/s2. BP a 2 r 7. From point z, draw vector zb' perpendicular to P2B to represent the tangential component of the acceleration of B with respect to P2 ( i.e. ) . a t BP 2 8. The vectors yb' and zb' intersect at b'. Now the vector p2' b' represents the acceleration of B with respect to P2 or the acceleration of B (i.e. ) or By measurement, we find that. 2 BP a B a aBP2 = aB = vector p2' b' = 29.6 m/s2 Ans. Also a a t BP vector yb = = 13.6 m/s2 , and vector zb = = 26.6 m/s2. t BA 2 We know that angular acceleration of P2B, 26 B P 6 . a t B = = = (Anticlockwise) Ans. 2 8 . 73 36 . 0 / rad s 2 2 P P B 2 and angular acceleration of AB, 6 . 13 AB a t BA (Anticlockwise) Ans. / 8 . 37 36 . 0 = = = rad s 2 AB

17. Ex(3) A mechanism of a crank and slotted lever quick return motion is shown in Figure (12). If the crank rotates counter clockwise at 120 r.p.m., determine for the configuration shown, the velocity and acceleration of the ram D. Also determine the angular acceleration of the slotted lever. Crank, AB = 150 mm ; Slotted arm, OC = 700 mm and link CD = 200 mm. Solution. Given : NBA = 120 r.p.m or BA = 2 120/60 = 12.57 rad/s ; AB = 150 mm = 0.15 m; OC = 700 mm = 0.7 m; CD = 200 mm = 0.2 m. = = 9 . 1 = 12 57 . . 0 15 / v AB m s We know that velocity of B with respect to A , BA BA ...(Perpendicular to AB) Draw the space diagram, to some suitable scale, as shown in Figure(a) the velocity diagram, as shown in Figure(b), is drawn as discussed below: Since O and A are fixed points, therefore these points are marked as one point in velocity diagram. Now draw vector ab in a direction perpendicular to AB , to some suitable scale, to represent the velocity of slider B with respect to A i.e.vBA, such that vector ab = vBA = 1.9 m/s. 1. From point o, draw vector ob' perpendicular to OB' to represent the velocity of coincident point B' (on the link OC) with respect to O i.e. vBO and from point b draw vector bb' parallel to the path of motion of B' (which is along the link OC) to represent the velocity of coincident point B' with respect to the slider B i.e. vB'B. The vectors ob' and bb' intersect at b'. 2. Figure (12) Since the point C lies on OB' produced, therefore, divide vector ob' at c in the same ratio as C divides OB' in the space diagram. In other words. ob / oc = OB /OC. The vector oc represents the velocity of C with respect to O i.e. vCO. 3. Now from point c, draw vector cd perpendicular to CD to represent the velocity of D with respect to C i.e. vDC and from point o draw vector od parallel to the path of motion of D (which is along the horizontal) to represent the velocity of D i.e. vD, The vectors cd and od intersect at d. 4.

18. By measurement, we find that velocity of the ram D, vD = vector od = 2.15 m/s Ans. From velocity diagram, we also find that Velocity of B with respect to B', vBB' = vector b'b = 1.05 m/s. Velocity of D with respect to C, vDC = vector cd = 0.45 m/s. Velocity of B' with respect to O, vB O = vector ob' = 1.55 m/s. Velocity of C with respect to O, respect to O, vCO = vector oc = 2.15 m/s. . 2 15 vCO Angular velocity of the link OC or OB', (Anticlockwise) O B CO = = = = . 3 07 / rad s 7 . 0 OC Acceleration of the ram D ( ) = = = 2 12 57 . . 0 15 23 7 . / We know that radial component of the acceleration of B with respect to A, AB rad s 2 ba 2 R BA Coriolis component of the acceleration of slider B with respect to the coincident point B', 05 . 1 07 . 3 2 2 2 v v a B B CO B B = = = = . 6 45 / m s 2 C ( . 0 ) 2 45 v 2 DC = = . 1 = 01 / a m s 2 r DC Radial component of the acceleration of D with respect to C, 2 . 0 CD Radial component of the acceleration of the coincident point B' with respect to O ( ) 2 / 62 . 4 52 . 0 O B 2 . 1 55 v 2 B = = = a m s r B O O

19. the acceleration diagram, as shown in Figure(d), is drawn as discussed below: 1.Since O and A are fixed points, therefore these points are marked as one point in the acceleration diagram. Draw vector a'b' parallel to AB, to some suitable scale, to represent the radial component of the acceleration of B with respect to A (i.e. Or aB, such that vector a b = = aB = 23.7 m/s . BA a BA a r r 2.The acceleration of the slider B with respect to the coincident point B' has the following two components : B a c i. Coriolis component of the acceleration of B with respect to B . i.e. . and B B a r ii.Radial component of the acceleration of B with respect to B' i.e. . B These two components are mutually perpendicular. Therefore from point b' draw vector b'x perpendicular to B'O i.e. in a direction as shown in Figure(c) to represent = 6.45 m/s2. The direction of is obtained by rotating vBB (represented by vector b'b in velocity diagram) through 90 in the same sense as that of link OC which rotates in the counter clockwise direction. Now from point x, draw vector xb'' perpendicular to vector b'x (or parallel to B'O) to represent whose magnitude is yet unknown . B B a B a r c B B B a r 3. The acceleration of the coincident point B' with respect to O has also the following two components: Ba i. Radial component of the acceleration of coincident point B' with respect to O , i.e. , and, r O Ba t ii. Tangential component of the acceleration of coincident point B' with respect to O, i.e. O These two components are mutually perpendicular. Therefore from point o', draw vector o'y parallel to B'O to represent =4.62 m/s2 and from point y draw vector yb'' perpendicular to vector o'y to represent The vectors xb'' and yb'' intersect at b''. Join o'b''. The vector o'b'' represents the acceleration of B' with respect to O . i.e. . Ba Ba r t O O B a O 4.Since the point C lies on OB' produced, therefore divide vector o'b'' at c' in the same ratio as C divides OB' in the space diagram. In other words o'b''/o'c' = OB'/OC .

20. 5. The acceleration of the ram D with respect to C has also the following two components: a i. Radial component of the acceleration of D with respect to C i.e. , and r DC a t DC ii.Tangential component of the acceleration of coincident point B' with respect to O, . a r DC The two components are mutually perpendicular. Therefore draw vector c'z parallel to CD to represent = 1.01 m/s2 and from z draw zd' perpendicular to vector zc' to represent whose magnitude is yet unknown . a t DC 6. From point o', draw vector o'd' in the direction of motion of the ram D which is along the horizontal. The vectors zd' and o'd' intersect at d'. The vector o'd' represents the acceleration of ram D, i.e . By measurement, we find that acceleration of the ram D, aD = vector o'd' = 8.4 m/s2 Ans. a DC Angular acceleration of the slotted lever t O Ba By measurement from acceleration diagram, we find that tangential component of the coincident point B' with respect to O, = vector yb = 6.4 m/s2. = 4 . 6 a t B = = 12 3 . / rad s 2 O O . 0 52 B We know that angular acceleration of the slotted lever, (Anticlockwise) Ans.

22. Exercises Q1/ In the mechanism, as shown in Figure(Q1), the crank OA rotates at 20 r.p.m. anticlockwise and gives motion to the sliding blocks B and D. The dimensions of the various links are OA = 300 mm; AB = 1200 mm; BC = 450 mm and CD = 450 mm. Figure (Q1) Q2/ Find out the acceleration of the slider D and the angular acceleration of link CD for the engine mechanism shown in Figure(Q2). The crank OA rotates uniformly at 180 r.p.m. in clockwise direction. The various lengths are: OA = 150 mm ; AB = 450 mm; PB = 240 mm ; BC = 210 mm ; CD = 660 mm. Figure (Q2)

23. Q3/ In a swiveling joint mechanism, as shown in Figure(Q3), the driving crank OA is rotating clockwise at 100 r.p.m. The lengths of various links are : OA = 50 mm ; AB = 350 mm; AD = DB ; DE = EF = 250 mm and CB = 125 mm. The horizontal distance between the fixed points O and C is 300 mm and the vertical distance between F and C is 250 mm. For the given configuration, determine: 1. Velocity of the slider block F, 2. Angular velocity of the link DE, 3. Velocity of sliding of the link DE in the swivel block, and 4. Acceleration of sliding of the link DE in the trunnion. Figure (Q3)