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- 1. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM LINEAR VELOCITY : The rate of change of linear displacement of body w.r.t. time is known as linear velocity. It is expressed by V or U. Linear velocity V = ds/dt It the displacement is along the circular path, the direction of velocity at any instant is along the tangent at that point. Fig. 2.1 LINEAR ACCELERATION : Rate of change of velocity w.r.t. time is linear acceleration. It is expressed by a. Linear acceleration a = dv/dt = d(ds/dt) dt = d2 s/dt2 Linear acceleration is also expressed as a = dv = ds x dv dt dt ds = V x dv ds ANGULAR DISPLACEMENT : It is defined as the angle described by a particle from one point to another w.r.t. time. As shown in fig. 2.2, when one line making angle 01 with x-axis moves through angle 02 with x-axis during short interval of time t2-t1 & making new position. Than 02-01 is known as the angular displacement of line. Fig. 2.2 Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi
- 2. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM ANGULAR VELOCITY : It is defined as the rate of change of angular displacement w.r.t. time. It is expressed by w. Angular velocity = d0/dt Note : If the direction of angular displacement is constant, it is known as angular speed. ANGULAR ACCELERATION : Rate of change of angular velocity w.r.t. time is defined as angular acceleration. It is expressed by o . Angular acceleration = dw = d(d0) = d2 0 dt dt(dt) dt2 RELATION BETWEEN ANGULAR VELOCITY & ROTATIONAL SPEED : If a body is rotating at the rate of N r.p.m., then the relation between angular speed & rotational speed is given as w = 2 II N rad/sec. 60 RELATION BETWEEN LINEAR & ANGULAR MOTION : Consider a body moving along a circular path from A to B as shown in fig. 2.3. Let, r = Radius of circular path 0 = Angular displacement in radians S = Linear displacement w = Angular velocity a = Linear acceleration o = Angular acceleration Fig. 2.3 From geometry of fig. 2.3, it is know that S = r 0 V = ds = d(r0) = r d0 = r . w ----------------------- (2.1) dt dt dt Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi
- 3. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM a = dv = d(rw) = r dw = r . o ---------------------- (2.2) dt dt dt MOTION OF PARTICLE ALONG A CIRCULAR PATH : Fig. 2.4 Consider P & Q, the two positions of particle displaced through an angle d0 in time dt. Now, V = Velocity of particle at P r = Radius of circular path V+dV = Velocity of particle at Q The change of velocity as the particle moves from P to Q obtained by drawing vector triangle OPQ as shown in fig. 2.4. Here OP represent velocity V & OQ represent velocity V+dV. The change of velocity in time dt is represented by PQ. a) Tangential component of acceleration : Velocity of P perpendicular to OP = V Velocity of Q perpendicular to OP = (V+dV)cosd0 Change of velocity = (V+dV)cosd0 – V Acceleration of P perpendicular to OP = (V+dV)cosd0 – V dt In the limit, as dt >0, cosd0 >1, So, tangential acceleration of component at = dv/dt = o . r ----------------------- (2.3) b) Centripetal or radial component of acceleration : Velocity of P parallel to OP = 0 Velocity of Q parallel to OP = (V+dV)sind0 Change of velocity = (V+dV)sind0 – 0 Acceleration of P parallel to OP = (V+dV)sind0 dt In the limit, as dt >0, sind0 >d0 So, centripetal/radial acceleration of component ar = V.d0/dt = V.w = (w.r).w = w2 .r ----------------------- (2.4) Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi P Q V r V+dVCOSd0 SINd0d0 d0
- 4. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM RELATIVE VELOCITY : To find relative motion between two bodies moving parallel or making some angle with each other, vector method is used. Va A Vb B Va Vb o a b Vab Vba Fig. 2.5 As shown in fig. 2.5. body A & B both are moving parallel to each other in same direction. Body A moves with velocity Va & B moves with velocity Vb. Here Va>Vb. Now relative velocity of A w.r.t. B = Vab = Va – Vb or ba = oa – ob. Similarly, Vba = Vb – Va or ab = ob – oa. As shown above relative velocity of A w.r.t. B is Vab & relative velocity of B w.r.t. A is Vba. Magnitude of both Vab & Vba will be same but direction will be different i.e. Vab = -Vba. VELOCITY & ACCELERATION IN SLIDER CRANK MECHANISM : As shown in fig. 2.6 slider C is connected to connecting rod BC & connecting rod BC is connected to crank AB at the point B. Let the crank radius is r, rotating about point A with uniform angular velocity w rad/sec in clockwise direction. Slider C reciprocate along the line of stroke AD. Here the magnitude & direction of velocity of B (Vb) is known. Fig. 2.6 Velocity : Using relative velocity method, velocity of slider C (Vc) can be determined. As shown in fig. 2.6(a), the velocity diagram is drawn by following step by step procedure. Vac a,d C Ve Vab e Vbc b Fig. 2.6(a) : Velocity Diagram Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi E
- 5. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM 1) From any point A draw the ab perpendicular to AB, such that ab = Vba = w.r to some suitable scale. 2) Link BC is rigid link hence velocity of C relative to B (Vcb) is perpendicular to BC. Now draw vector bc perpendicular to BC to represent Vcb. 3) From point A, draw vector parallel to the path of motion of slider C. The vector bc & ac intersect at c. Now ac represents the velocity of slider C relative to A (Vca). 4) Velocity of any point E on the connecting rod can be obtained by dividing vector bc in appropriate proportionate. i.e. be/bc = BE/BC. To get point e, draw the line ae from point a to point e on bc. 5) Find angular velocity of BC, wcb = Vcb/CB. Acceleration : The crank AB rotates at a uniform velocity, so the acceleration of B w.r.t. A has only radial/centripetal component. Similarly, the slider moves in a linear direction, thus slider C has no radial/centripetal component. As shown in fig. 2.6(b) the acceleration diagram is drawn by following step by step procedure. c aac a ae at bc e aab m ar bc b Fig. 2.6(b) Acceleration Diagram 1) The radial/centripetal component of B w.r.t. A or acceleration of B w.r.t. A = aba = (Vba)2 /AB. (Here a point at the end of link which moves with constant angular velocity has no tangential component, so tangential component is zero). Draw line ab parallel to AB equal to aba with some suitable scale. 2) From point b draw line bm parallel to BC which represent radial component of slider C w.r.t. B. which can find by ar cb = (Vcb)2 /CB. 3) From point m draw line mc perpendicular to line bm. The vector mc represent the tangential component of acceleration of C w.r.t. B i.e. at cb. (Here a point moves along straight line has no radial component of acceleration). 4) Draw line ac parallel to AC, which intersect the vector mc at c. 5) The vector bc which is the sum of vector bm & mc represents the total acceleration of C w.r.t. B. i.e. acb, acceleration of connecting rod BC. 6) The acceleration of any point on connecting rod BC such as E may be obtained by dividing vector bc in appropriate proportionate. i.e. be/bc = BE/BC. 7) Find angular acceleration of connecting rod BC = o cb = at cb/CB. VELOCITY & ACCELERATION IN SLIDER CRANK MECHANISM : As shown in fig. 2.7 ABCD show four bar chain mechanism in which AD is the fixed link & BC is coupler. AB is the driver rotating at an angular speed of w rad/sec in the clockwise direction. Here the magnitude & direction of velocity of B (Vb) is known. Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi
- 6. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM Fig. 2.7 Velocity : The velocity diagram for the four bar chain as shown in fig. 2.7(a) is drawn as per below procedure. c Vdc a,d e Vbc Vab b Fig. 2.7(a) Velocity Diagram 1) As the link AD is fixed, the point a & d are taken as one point in the velocity diagram. From any point a draw the ab perpendicular to AB, such that ab = Vba = w.r to some suitable scale. 2) Now from point b draw vector bc perpendicular to link BC to represent the velocity of C w.r.t. B (Vcb). 3) From point a,d draw vector dc perpendicular to link CD to represent the velocity of C w.r.t. D (Vcd). The vector bc & dc intersect at c. 4) Velocity of any point E on the connecting rod can be obtained by dividing vector bc in appropriate proportionate. i.e. be/bc = BE/BC. To get point e, draw the line ae from point a to point e on bc. 5) Find angular velocity of BC, wcb = Vcb/CB. Acceleration : Since the angular velocity of AB is uniform, so there will be no tangential component of the acceleration of B w.r.t. A. The radial component of acceleration of B w.r.t. A = ar ba = aba = (Vba)2 /BA. a,d adc ar dc aba c y e at dc at bc acb b x ar bc Fig. 2.7(b) Acceleration Diagram Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi
- 7. CHAPTER – 2 : VELOCITY & ACCELERATION DIAGRAM The radial component of acceleration of C w.r.t. B = ar cb = (Vcb)2 /CB. The radial component of acceleration of C w.r.t. D = ar cd = (Vcd)2 /CD. Now acceleration diagram as shown in fig. 2.7(b) is drawn as follows. 1) Draw the vector ab parallel to AB with some suitable scale to represent the radial component of the acceleration of B w.r.t. A, such that vector ab = ar ba = aba. 2) From point b draw vector bx parallel to BC to represent the radial component of the acceleration of C w.r.t. B, such that vector bx = ar cb. 3) From point x draw xc perpendicular to bx to represent the tangential component of C w.r.t. B i.e. at cb, whose magnitude is not yet taken. 4) From point d draw dy parallel to CD to represent the radial component of C w.r.t. D i.e. vector dy = ar cd. 5) From point y draw yc perpendicular to dy to represent the tangential component of C w.r.t. D i.e. at cd. 6) The vector xc & yc intersect at c. Join ca & cb. 7) The acceleration of any point on connecting rod BC such as E may be obtained by dividing vector bc in appropriate proportionate. i.e. be/bc = BE/BC. 8) Find angular acceleration of CD = o cd = at cd/cd. ~*~*~*~*~*~*~*~*~*~*~*~*~*~*~ Theory of Machine (1MEB25) Prepared By : Prof. Jignesh Dangi

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