Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Beer Dinamica 8e Presentaciones Cap... by Jorge Enrique Men... 7225 views
- R05010105 A P P L I E D M E C H A... by guestd436758 1945 views
- Beer vector mechanics for engineers... by Mario670m 244093 views
- Kinematika partikel by badriyatul 8072 views
- Xi bab kinematika vektor marthen by eli priyatna laidan 157 views
- Es62finalexam by Maylanie Sarah 76 views

1,198 views

979 views

979 views

Published on

kinematika partikel

No Downloads

Total views

1,198

On SlideShare

0

From Embeds

0

Number of Embeds

0

Shares

0

Downloads

72

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Kinematics of Particles
- 2. ContentsProblem 11.5 Sample Introduction Rectilinear Motion: Position, Velocity & Acceleration Graphical Solution of RectilinearDetermination of the Motion of a ParticleMotion Problems Sample Problem 11.2 Other Graphical Methods Sample Problem 11.3 Curvilinear Motion: Position, Velocity Uniform Rectilinear-Motion & Acceleration Uniformly Accelerated RectilinearDerivatives of Vector Functions Motion Rectangular Components of Velocity Motion of Several Particles: and Acceleration Relative Motion Motion Relative to a Frame in Sample Problem 11.4 Translation Motion of Several Particles: Tangential and Normal Components Dependent Motion Radial and Transverse Components Sample Problem 11.10 Sample Problem 11.12 11 - 2
- 3. • Dynamics includes: Introduction - Kinematics: study of the geometry of motion. Kinematics is used to relate displacement, velocity, acceleration, and time without reference to the cause of motion. - Kinetics: study of the relations existing between the forces acting on a body, the mass of the body, and the motion of the body. Kinetics is used to predict the motion caused by given forces or to determine the forces required to produce a given motion. • Rectilinear motion: position, velocity, and acceleration of a particle as it moves along a straight line. • Curvilinear motion: position, velocity, and acceleration of a particle as it moves along a curved line in two or three dimensions. 11 - 3
- 4. Rectilinear Motion: Position, Velocity & Acceleration • Particle moving along a straight line is said to be in rectilinear motion. • Position coordinate of a particle is defined by positive or negative distance of particle from a fixed origin on the line. • The motion of a particle is known if the position coordinate for particle is known for every value of time t. Motion of the particle may be expressed in the form of a function, e.g., x = 6t 2 − t 3 or in the form of a graph x vs. t. 11 - 4
- 5. Rectilinear Motion: Position, Velocity & Acceleration • Consider particle which occupies position P at time t and P’ at t+∆t, ∆x = Average velocity ∆t ∆x = v = lim Instantaneous velocity ∆t →0 ∆t • Instantaneous velocity may be positive or negative. Magnitude of velocity is referred to as particle speed. • From the definition of a derivative, ∆x dx v = lim = dt ∆t →0 ∆t e.g., x = 6t 2 − t 3 dx v= = 12t − 3t 2 dt 11 - 5
- 6. Rectilinear Motion: Position, Velocity & Acceleration • Consider particle with velocity v at time t and v’ at t+∆t, ∆v Instantaneous acceleration = a = lim ∆t →0 ∆t • Instantaneous acceleration may be: - positive: increasing positive velocity or decreasing negative velocity - negative: decreasing positive velocity or increasing negative velocity. • From the definition of a derivative, ∆v dv d 2 x a = lim = = 2 dt dt ∆t →0 ∆t e.g. v = 12t − 3t 2 dv a= = 12 − 6t dt 11 - 6
- 7. Rectilinear Motion: Position, Velocity & Acceleration motion given by • Consider particle with x = 6t 2 − t 3 v= dx = 12t − 3t 2 dt dv d 2 x a= = = 12 − 6t dt dt 2 • at t = 0, x = 0, v = 0, a = 12 m/s2 • at t = 2 s, x = 16 m, v = vmax = 12 m/s, a = 0 • at t = 4 s, x = xmax = 32 m, v = 0, a = -12 m/s2 • at t = 6 s, x = 0, v = -36 m/s, a = 24 m/s2 11 - 7
- 8. Determination of the Motion of a Particle • Recall, motion of a particle is known if position is known for all time t. • Typically, conditions of motion are specified by the type of acceleration experienced by the particle. Determination of velocity and position requires two successive integrations. • Three classes of motion may be defined for: - acceleration given as a function of time, a = f(t) - acceleration given as a function of position, a = f(x) - acceleration given as a function of velocity, a = f(v) 11 - 8
- 9. Determination of the Motion of a Particle • Acceleration given as a function of time, a = f(t): v( t ) t dv = a = f (t) dv = f ( t ) dt ∫ dv = ∫ f ( t ) dt dt v 0 dx = v( t ) dt x( t ) dx = v( t ) dt t v( t ) − v0 = ∫ f ( t ) dt 0 0 t t ∫ dx = ∫ v( t ) dt x0 x( t ) − x0 = ∫ v( t ) dt 0 0 • Acceleration given as a function of position, a = f(x): v= dx dx or dt = dt v v dv = f ( x ) dx a= v( x ) dv dv or a = v = f ( x ) dt dx x ∫ v dv = ∫ f ( x ) dx v0 x0 1 v( x ) 2 2 2 − 1 v0 2 = x ∫ f ( x ) dx x0 11 - 9
- 10. Determination of the Motion of a Particle • Acceleration given as a function of velocity, a = f(v): dv = a = f ( v) dt v( t ) ∫ v0 dv = dt f ( v) v( t ) ∫ v0 t dv = ∫ dt f ( v) 0 dv =t f ( v) dv v = a = f ( v) dx x ( t ) − x0 = v( t ) ∫ v0 v dv dx = f ( v) x( t ) v( t ) x0 v0 ∫ dx = ∫ v dv f ( v) v dv f ( v) 11 - 10
- 11. Sample Problem 11.2 SOLUTION: • Integrate twice to find v(t) and y(t). • Solve for t at which velocity equals zero (time for maximum elevation) and evaluate corresponding altitude. Ball tossed with 10 m/s vertical velocity from window 20 m above ground. Determine: • velocity and elevation above ground at time t, • highest elevation reached by ball and corresponding time, and • time when ball will hit the ground and corresponding velocity. • Solve for t at which altitude equals zero (time for ground impact) and evaluate corresponding velocity. 11 - 11
- 12. Sample Problem 11.2 SOLUTION: • Integrate twice to find v(t) and y(t). dv = a = −9.81 m s 2 dt v( t ) t v( t ) − v0 = −9.81t ∫ dv = − ∫ 9.81 dt v0 0 v( t ) = 10 dy = v = 10 − 9.81t dt y( t ) t ∫ dy = ∫ (10 − 9.81t ) dt y0 0 m m − 9.81 2 t s s y ( t ) − y0 = 10t − 1 9.81t 2 2 m m y ( t ) = 20 m + 10 t − 4.905 2 t 2 s s 11 - 12
- 13. Sample Problem 11.2 zero and evaluate • Solve for t at which velocity equals corresponding altitude. v( t ) = 10 m m − 9.81 2 t = 0 s s t = 1.019 s • Solve for t at which altitude equals zero and evaluate corresponding velocity. m m y ( t ) = 20 m + 10 t − 4.905 2 t 2 s s m m y = 20 m + 10 (1.019 s ) − 4.905 2 (1.019 s ) 2 s s y = 25.1 m 11 - 13
- 14. Sample Problem 11.2 • Solve for t at which altitude equals zero and evaluate corresponding velocity. m m y ( t ) = 20 m + 10 t − 4.905 2 t 2 = 0 s s t = −1.243 s ( meaningless ) t = 3.28 s v( t ) = 10 m m − 9.81 2 t s s v( 3.28 s ) = 10 m m − 9.81 2 ( 3.28 s ) s s v = −22.2 m s 11 - 14
- 15. Sample Problem 11.3 SOLUTION: a = − kv • Integrate a = dv/dt = -kv to find v(t). • Integrate v(t) = dx/dt to find x(t). Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in fixed cylinder filled with oil. As barrel recoils with initial velocity v0, piston moves and oil is forced through orifices in piston, causing piston and cylinder to decelerate at rate proportional to their velocity. • Integrate a = v dv/dx = -kv to find v(x). Determine v(t), x(t), and v(x). 11 - 15
- 16. Sample Problem 11.3 SOLUTION: • Integrate a = dv/dt = -kv to find v(t). v( t ) t dv dv v( t ) a= = − kv = − k ∫ dt ln = − kt ∫ v dt v0 v 0 0 v( t ) = v0 e − kt • Integrate v(t) = dx/dt to find x(t). dx v( t ) = = v0 e − kt dt t x( t ) t 1 − kt − kt x ( t ) = v0 − e ∫ dx = v0 ∫ e dt k 0 0 0 x( t ) = ( v0 1 − e − kt k 11 - 16 )
- 17. Sample Problem 11.3 • Integrate a = v dv/dx = -kv to find v(x). v x v0 dv a = v = − kv dx 0 dv = − k dx ∫ dv = −k ∫ dx v − v0 = − kx v = v0 − kx • Alternatively, ( v0 1 − e − kt k ) with x( t ) = and v( t ) = v0 e − kt or e − kt = then x( t ) = v0 v ( t ) 1 − k v0 v( t ) v0 v = v0 − kx 11 - 17
- 18. Uniform Rectilinear Motion For particle in uniform rectilinear motion, the acceleration is zero and the velocity is constant. dx = v = constant dt x t x0 0 ∫ dx = v ∫ dt x − x0 = vt x = x0 + vt 11 - 18
- 19. Uniformly Accelerated Rectilinear Motion For particle in uniformly accelerated rectilinear motion, the acceleration of the particle is constant. v t v0 dv = a = constant dt 0 ∫ dv = a ∫ dt v − v0 = at v = v0 + at dx = v0 + at dt x t x0 0 ∫ dx = ∫ ( v0 + at ) dt x − x0 = v0 t + 1 at 2 2 x = x0 + v0t + 1 at 2 2 dv v = a = constant dx v x v0 x0 ∫ v dv = a ∫ dx 1 2 (v 2 − v02 ) = a( x − x0 ) 2 v 2 = v0 + 2 a ( x − x0 ) 11 - 19
- 20. Motion of Several Particles: Relative Motion • For particles moving along the same line, time should be recorded from the same starting instant and displacements should be measured from the same origin in the same direction. = x B − x A = relative position of B with respect to A xB = x A + xB A xB A = v B − v A = relative velocity of B with respect to A vB = v A + vB A vB A = a B − a A = relative acceleration of B with respect to A aB = a A + aB A aB A 11 - 20
- 21. Sample Problem 11.4 SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. Ball thrown vertically from 12 m level in elevator shaft with initial velocity of 18 m/s. At same instant, open-platform elevator passes 5 m level moving upward at 2 m/s. Determine (a) when and where ball hits elevator and (b) relative velocity of ball and elevator at contact. • Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. • Substitute impact time into equation for position of elevator and relative velocity of ball with respect to elevator. 11 - 21
- 22. Sample Problem 11.4 SOLUTION: • Substitute initial position and velocity and constant acceleration of ball into general equations for uniformly accelerated rectilinear motion. v B = v0 + at = 18 m m − 9.81 2 t s s m m y B = y0 + v0 t + 1 at 2 = 12 m + 18 t − 4.905 2 t 2 2 s s • Substitute initial position and constant velocity of elevator into equation for uniform rectilinear motion. m vE = 2 s m y E = y0 + v E t = 5 m + 2 t s 11 - 22
- 23. Sample Problem 11.4 • Write equation for relative position of ball with respect to elevator and solve for zero relative position, i.e., impact. yB E ( ) = 12 + 18t − 4.905t 2 − ( 5 + 2t ) = 0 t = −0.39 s ( meaningless ) t = 3.65 s • Substitute impact time into equations for position of elevator and relative velocity of ball with respect to elevator. y E = 5 + 2( 3.65) y E = 12.3 m vB E = (18 − 9.81t ) − 2 = 16 − 9.81( 3.65) vB E = −19.81 m s 11 - 23
- 24. Motion of Several Particles: Dependent Motion on position of one • Position of a particle may depend or more other particles. • Position of block B depends on position of block A. Since rope is of constant length, it follows that sum of lengths of segments must be constant. x A + 2 x B = constant (one degree of freedom) • Positions of three blocks are dependent. 2 x A + 2 x B + xC = constant (two degrees of freedom) • For linearly related positions, similar relations hold between velocities and accelerations. dx dx A dx + 2 B + C = 0 or 2v A + 2v B + vC = 0 dt dt dt dv dv dv 2 A + 2 B + C = 0 or 2a A + 2a B + aC = 0 dt dt dt 2 11 - 24
- 25. Sample Problem 11.5 SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. • Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L. • Pulley D has uniform rectilinear motion. Pulley D is attached to a collar which Calculate change of position at time t. is pulled down at 3 in./s. At t = 0, collar A starts moving down from K • Block B motion is dependent on motions of collar A and pulley D. Write motion with constant acceleration and zero relationship and solve for change of block initial velocity. Knowing that B position at time t. velocity of collar A is 12 in./s as it passes L, determine the change in • Differentiate motion relation twice to elevation, velocity, and acceleration develop equations for velocity and of block B when block A is at L. acceleration of block B. 11 - 25
- 26. Sample Problem 11.5 SOLUTION: • Define origin at upper horizontal surface with positive displacement downward. • Collar A has uniformly accelerated rectilinear motion. Solve for acceleration and time t to reach L. 2 v 2 = ( v A ) 0 + 2a A [ x A − ( x A ) 0 ] A 2 in. 12 = 2a A ( 8 in.) s aA = 9 in. s2 v A = ( v A ) 0 + a At 12 in. in. =9 2t s s t = 1.333 s 11 - 26
- 27. Sample Problem rectilinear motion. Calculate 11.5 • Pulley D has uniform change of position at time t. xD = ( xD ) 0 + vDt in. x D − ( x D ) 0 = 3 (1.333 s ) = 4 in. s • Block B motion is dependent on motions of collar A and pulley D. Write motion relationship and solve for change of block B position at time t. Total length of cable remains constant, x A + 2 x D + x B = ( x A ) 0 + 2( x D ) 0 + ( x B ) 0 [ x A − ( x A ) 0 ] + 2[ xD − ( x D ) 0 ] + [ x B − ( x B ) 0 ] = 0 ( 8 in.) + 2( 4 in.) + [ x B − ( x B ) 0 ] = 0 x B − ( x B ) 0 = −16 in. 11 - 27
- 28. Sample Problem 11.5 to develop • Differentiate motion relation twice equations for velocity and acceleration of block B. x A + 2 x D + x B = constant v A + 2v D + v B = 0 in. in. 12 + 2 3 + v B = 0 s s v B = 18 in. s a A + 2a D + a B = 0 in. 9 2 + vB = 0 s a B = −9 in. s2 11 - 28
- 29. Graphical Solution of RectilinearMotion Problems • Given the x-t curve, the v-t curve is equal to the x-t curve slope. • Given the v-t curve, the a-t curve is equal to the v-t curve slope. 11 - 29
- 30. Graphical Solution of RectilinearMotion Problems • Given the a-t curve, the change in velocity between t1 and t2 is equal to the area under the a-t curve between t1 and t2. • Given the v-t curve, the change in position between t1 and t2 is equal to the area under the v-t curve between t1 and t2. 11 - 30
- 31. Other Graphical Methods • Moment-area method to determine particle position at time t directly from the a-t curve: x1 − x0 = area under v − t curve = v0t1 + v1 ∫ ( t1 − t ) dv v0 using dv = a dt , x1 − x0 = v0t1 + v1 ∫ ( t1 − t ) a dt v0 v1 ∫ ( t1 − t ) a dt = first moment of area under a-t curve v0 with respect to t = t1 line. x1 = x0 + v0t1 + ( area under a-t curve)( t1 − t ) t = abscissa of centroid C 11 - 31
- 32. Other Graphical Methods • Method to determine particle acceleration from v-x curve: dv a=v dx = AB tan θ = BC = subnormal to v-x curve 11 - 32
- 33. Curvilinear Motion: Position, • Particle Acceleration Velocity &moving along a curve other than a straight line is in curvilinear motion. • Position vector of a particle at time t is defined by a vector between origin O of a fixed reference frame and the position occupied by particle. • Consider particle which occupies position P r′ defined by at time t and P’ defined by at t + r ∆t, ∆r dr v = lim = dt ∆t →0 ∆t = instantaneous velocity (vector) ∆s ds = dt ∆t →0 ∆t v = lim = instantaneous speed (scalar) 11 - 33
- 34. Curvilinear Motion: Position, • Consider velocity v of particle at time t and velocity Velocity v& Acceleration ′ at t + ∆t, ∆v dv a = lim = dt ∆t →0 ∆t = instantaneous acceleration (vector) • In general, acceleration vector is not tangent to particle path and velocity vector. 11 - 34
- 35. Derivatives P( u ) be a vector function of scalar variable u, of Vector Functions • Let dP ∆P P( u + ∆u ) − P ( u ) = lim = lim du ∆u →0 ∆u ∆u →0 ∆u • Derivative of vector sum, d ( P + Q ) dP dQ = + du du du • Derivative of product of scalar and vector functions, d ( f P ) df dP = P+ f du du du • Derivative of scalar product and vector product, d ( P • Q ) dP dQ = •Q + P• du du du dQ d ( P × Q ) dP = ×Q + P× du du du 11 - 35
- 36. Rectangular Components of • & Acceleration VelocityWhen position vector of particle P is given by its rectangular components, r = xi + y j + zk • Velocity vector, dx dy dz v = i + j + k = xi + y j + zk dt dt dt = vx i + v y j + vz k • Acceleration vector, d 2 x d 2 y d 2 z a = 2 i + 2 j + 2 k = i + j + k x y z dt dt dt = ax i + a y j + az k 11 - 36
- 37. Rectangular Components of • Rectangular components particularly Velocity &component accelerations can beeffective Acceleration integrated when independently, e.g., motion of a projectile, a x = = 0 x a y = = − g y a z = = 0 z with initial conditions, ( vx ) 0 , v y , ( vz ) 0 = 0 x0 = y 0 = z 0 = 0 ( )0 Integrating twice yields vx = ( vx ) 0 x = ( vx ) 0 t ( ) 0 − gt y = ( v y ) y − 1 gt 2 2 0 vy = vy vz = 0 z=0 • Motion in horizontal direction is uniform. • Motion in vertical direction is uniformly accelerated. • Motion of projectile could be replaced by two independent rectilinear motions. 11 - 37
- 38. Motion Relative to a Frame in • Designate one frame as of Translation the fixed frame thereference. All other frames not rigidly attached to fixed reference frame are moving frames of reference. • Position vectors for particles A and B with respect to the fixed frame of reference Oxyz are rA and rB . rB A joining A and B defines the position of • Vector B with respect to the moving frame Ax’y’z’ and rB = rA + rB A • Differentiating twice, vB = v A + vB A vB a B = a A + aB A aB A A = velocity of B relative to A. = acceleration of B relative to A. • Absolute motion of B can be obtained by combining motion of A with relative motion of B with respect to moving reference frame attached to A. 11 - 38
- 39. Tangential and Normal Componentsparticle is tangent to path of • Velocity vector of particle. In general, acceleration vector is not. Wish to express acceleration vector in terms of tangential and normal components. • et and et′ are tangential unit vectors for the particle path at P and P’. When drawn with respect to the same origin, ∆et = et′ − et and ∆θ is the angle between them. ∆et = 2 sin ( ∆θ 2) ∆et sin ( ∆θ 2 ) lim = lim en = en ∆θ →0 ∆θ ∆θ →0 ∆θ 2 det en = dθ 11 - 39
- 40. Tangential and Normal • With the velocity vector expressed as v = ve Components may be written as the particle acceleration t de dv de dθ ds dv dv a= = et + v = et + v dt dt dt dt dθ ds dt but det ds = en ρ dθ = ds =v dθ dt After substituting, dv v2 dv v 2 a = et + en at = an = dt ρ dt ρ • Tangential component of acceleration reflects change of speed and normal component reflects change of direction. • Tangential component may be positive or negative. Normal component always points toward center of path curvature. 11 - 40
- 41. Tangential and Normal Components and normal acceleration • Relations for tangential also apply for particle moving along space curve. dv v 2 a = et + en dt ρ dv at = dt v2 an = ρ • Plane containing tangential and normal unit vectors is called the osculating plane. • Normal to the osculating plane is found from eb = et × en en = principal normal eb = binormal • Acceleration has no component along binormal. 11 - 41
- 42. Radial and Transverse Components • When particle position is given in polar coordinates, it is convenient to express velocity and acceleration with components parallel and perpendicular to OP. r = re r der = eθ dθ • The particle velocity vector is der dr dr dθ d v = ( r er ) = e r + r = er + r eθ dt dt dt dt dt = r er + rθ eθ deθ = − er dθ der der dθ dθ = = eθ dt dθ dt dt deθ deθ dθ dθ = = − er dt dθ dt dt • Similarly, the particle acceleration vector is dθ d dr a = er + r eθ dt dt dt d 2 r dr der dr dθ d 2θ dθ deθ = 2 er + + eθ + r 2 eθ + r dt dt dt dt dt dt dt dt = − rθ 2 er + ( rθ + 2rθ ) eθ r ( ) 11 - 42
- 43. Radial and Transverse position is given in cylindrical Components • When particle coordinates, it is convenient to express the velocity and acceleration vectors using the unit vectors eR , eθ , and k . • Position vector, r = R e R +z k • Velocity vector, dr eθ + z k v= = R e R + Rθ dt • Acceleration vector, dv 2 a= = R − Rθ e R + ( Rθ + 2 Rθ ) eθ + k z dt ( ) 11 - 43
- 44. Sample Problem 11.10 SOLUTION: • Calculate tangential and normal components of acceleration. • Determine acceleration magnitude and direction with respect to tangent to curve. A motorist is traveling on curved section of highway at 60 mph. The motorist applies brakes causing a constant deceleration rate. Knowing that after 8 s the speed has been reduced to 45 mph, determine the acceleration of the automobile immediately after the brakes are applied. 11 - 44
- 45. SampleSOLUTION: Problem 11.10 • Calculate tangential and normal components of acceleration. ∆v ( 66 − 88) ft s ft at = = = −2.75 2 ∆t 8s s v 2 ( 88 ft s ) 2 ft an = = = 3.10 2 ρ 2500 ft s 60 mph = 88 ft/s 45 mph = 66 ft/s • Determine acceleration magnitude and direction with respect to tangent to curve. ft 2 2 2 2 a = 4.14 2 a = at + a n = ( − 2.75) + 3.10 s α = tan −1 a n at = tan −1 3.10 2.75 α = 48.4° 11 - 45
- 46. Sample Problem 11.12 SOLUTION: • Evaluate time t for θ = 30o. • Evaluate radial and angular positions, and first and second derivatives at time t. Rotation of the arm about O is defined by θ = 0.15t2 where θ is in radians and t in seconds. Collar B slides along the arm such that r = 0.9 - 0.12t2 where r is in meters. • Calculate velocity and acceleration in cylindrical coordinates. • Evaluate acceleration with respect to arm. After the arm has rotated through 30o, determine (a) the total velocity of the collar, (b) the total acceleration of the collar, and (c) the relative acceleration of the collar with respect to the arm. 11 - 46
- 47. Sample Problem 11.12 SOLUTION: • Evaluate time t for θ = 30o. θ = 0.15 t 2 = 30° = 0.524 rad t = 1.869 s • Evaluate radial and angular positions, and first and second derivatives at time t. r = 0.9 − 0.12 t 2 = 0.481 m r = −0.24 t = −0.449 m s r = −0.24 m s 2 θ = 0.15 t 2 = 0.524 rad θ = 0.30 t = 0.561 rad s θ = 0.30 rad s 2 11 - 47
- 48. Sample Problem acceleration. 11.12 • Calculate velocity and vr = r = −0.449 m s vθ = rθ = ( 0.481m )( 0.561rad s ) = 0.270 m s v β = tan −1 θ vr 2 2 v = vr + vθ v = 0.524 m s β = 31.0° ar = − rθ 2 r = −0.240 m s 2 − ( 0.481m ) ( 0.561rad s ) 2 = −0.391m s 2 aθ = rθ + 2rθ ( ) = ( 0.481m ) 0.3 rad s 2 + 2( − 0.449 m s )( 0.561rad s ) = −0.359 m s 2 2 2 a = ar + aθ a γ = tan −1 θ ar a = 0.531m s γ = 42.6° 11 - 48
- 49. Sample •Problem 11.12 to arm. Evaluate acceleration with respect Motion of collar with respect to arm is rectilinear and defined by coordinate r. a B OA = r = −0.240 m s 2 11 - 49

No public clipboards found for this slide

Be the first to comment