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Dynamics

a branch of mechanics that deals with the “ motion ” of bodies under the action of force s. Dynamics. Two parts:. Kinematics Study of motion without reference to the forces that cause the motion. 2. Kinetics

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Dynamics

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  1. a branch of mechanics that deals with the “motion” of bodies under the action of forces. Dynamics Two parts: • Kinematics • Study of motion without reference to the forces that cause the motion 2. Kinetics Study of motion under the action of forces on bodies for resulting motions.

  2. Kinematics Kinetics study of motion under the action of forces on bodies about the bodies’ motions. study of objects’ motion without reference to the forces that cause the motion kinematics relation is necessary to solve kinetics problem How does wABrelated with another wＣＤ (kinematics) If you want aCD = 36.5 rad/s2, how much torques do you apply to AB (kinetics)

  3. Kinematics Kinetics Particles Before Midterm After Midterm Rigid Bodies

  4. How to describe the motion? • = “position vector” (start from some convenient point: reference point) Kinematics of particles How to describe the position? • Frame: Ref-Point + Coordinate Motion: time-related position Coordinate Path of the particle (x,y) coord time-related info: velocity , acceleration A' (r,q) coord • from A to A' takes t second s r A q • Distant traveled: measured along the path, scalar O Reference Point You are going to learn: • Displacement( in t ): r-qcoord (x,y) coord n-t coord relative frame

  5. 2/2 Rectilinear Motion (1D-motion) By defining the axis according to the moving direction “displacement, velocity, acceleration” can be considered as scalar quantity. Particle moving on straight linepath. reference point The particle is at point P at time t and point P at time t+t with a moving distance of s. P: t P’: t+Dt reference axis +s D s “average” velocity: Positive v is defined in the same direction as positive s. i.e.) positive v implies that s is increasing, and negative v implies that s is deceasing. “instantaneous” velocity :

  6. Rectilinear Motion “instantaneous” acceleration ( similarly as ) or • Eliminating dt, we have or 3 Equations, but only 2 are independent Positive a is defined in the same direction as positive v (or s). Ex. positive a implies that the particle is speeding up (accelerating) negative a implies that the particle is slowing down (decelerating).

  7. Formula Interpretations of a,v,s,t (1) Constant acceleration (a= constant).Find v(t), s(t)

  8. Ans 5 m

  9. a=g (const) Time required for 3-m falling. s2 s1 =3 2nd Thus, the 2nd ball has time to fall: 1st Therefore, the 2nd ball travels:

  10. v(t) s(t) Formula Interpretations of a,v,s,t (2) Acceleration given as a function of time, a = f(t) . Find v(t), s(t) Or by solving the differential equation: s(t) v(t)

  11. Formula Interpretations of a,v,s,t t(v) s(v) v(t) s(t) Find s(v), t(v) (3) Acceleration given as a function of velocity, a = f(v) Find v(t), s(t), a(t)

  12. v(s) t(s) s(t) a(t) v(t) Formula Interpretations of a,v,s,t Find s(t), v(t), a(t) (4) Acceleration as function of displacement: a = f (s)Find v(s), t(s)

  13. Graphical Interpretations of a, v, s, t Area under a-t curve from t1-t2 = v(t2) − v(t1) • The familiar slopes and areas s-t curve a-t curve v-t curve Area under v-t curve from t1-t2 = s(t2) − s(t1)

  14. Graphical Interpretations of a, v, s, t • Less familiar interpretations Find a ! v-s curve a-s curve Find v ! q q

  15. Given the acceleration-displacement plot as shown. Determine the velocity when x = 1.4 m, assuming that the velocity is 0.8 m/s at x = 0 a-s curve Area under curve = 0.16 + 0.12 + 0.08 + 0 = 0.36 1.4 + or -? ANS

  16. v0 v a=a(v) s Fig1 Case (b) Case (a)

  17. Find v(t), s(t) of the mass if s start at zero and + or - ? Altenative Solution: Differential equation s

  18. downward: Upward: Find h and v when the ball hits the ground. = 24.1 m/s = 36.5 m

  19. SP2/1 Given: x Find 1) t when v=+72 How many turns? 2) a when v=+32 3) total distance traveled from t=1 to t=4 Correct? total distance traveled net displacement != total distance traveled

  20. 2/44 The electronic trottle control of a model train is programmed so that the train speed varies with position as shown. Determine the time t required for the train to complete one lap. Rectilinear motion? a along the path not total a v, a v, a s s Rectilinear equations can be used for curved motion if s, v, a are measured along the curve (more on this soon)

  21. 2/44 The electronic trottle control of a model train is programmed so that the train speed varies with position as shown. Determine the time t required for the train to complete one lap. Area = vds Slope = dv/ds = Slope = Constant. =C = 50.8 s

  22. Recommended Problem 2/29 2/36 2/46 2/58 slope =0 at both end

  23. = “position vector” (start from some convenient point: reference point) Kinematics of particles How to describe this particle’s motion? • Framework for describing the motion Reference Frame Path of the particle (x,y) coord A' (r,q) coord • from A to A' takes t second s r A • displacement( in t ): q O Reference Point • Distant traveled: measured along the path, scalar relative frame You are going to learn: (x,y) coord n-t coord r-qcoord

  24. = “position vector” (start from some convenient point: reference point) 2/3 Plane Curvilinear Motion • Motion of a particle along a curved path in a single plane (2D curve) Reference Frame Path of the particle (x,y) coord • from A to A' takes t second A' • displacementof the particle during time t : (r,q) coord s r • Distant traveled = s, measured along the path, scalar A q O • Basic Concept: “time derivative of a vector” “Average Velocity” “(Instantaneous) Velocity”

  25. Speed and Velocity “(Instantaneous) Velocity” Path of the particle A' “(Instantaneous) Speed” s A O Path of the particle A O Velocity vector is tangent to the curve path

  26. Instantaneous Speed Path of the particle A' s A O Time rate of change of length of the position vector speed (r,q) coord r q

  27. Hodograph Hodograph C C “Average acceleration” “(Instantaneous) Acceleration” Acceleration Path of the particle A' A O *** The velocity is always tangent to the path of the particle (frequently used in problems) while the acceleration is tangent to the hodograph (not very important) ***

  28. Vector Equation and reference frame Path of the particle Reference Frame A' s A Vector equation is in general form, not depending on used coordinate. O • Rectangular: x-y • Normal-Tangent: n-t • Polar: r- Reference Frame (coordinate) Usage will depend on selection. More than one can be used At the same time. n-t rectangular r-q

  29. Derivatives of Vectors Derivatives of Vectors: Obey the same rules as they do for scalars

  30. Reference Frame 2/4 Rectangular Coordinates O Path Both “divided” particles, are moving in rectilinear motion y O O x Correct? Basic Agreement: Direction of reference axis {x,y} do not change on time variation. 0 0 A Rectilinear Motion in 2 perpendicular & independent axes.

  31. Path y y x O O x rectilinear in 2 dimension,related which other via time. If given can you find rectilinear in y-axis rectilinear in x-axis Velocity is tangent to the path (= slope of curve path)

  32. Common Cases Rectangular coordinates are usually good for problems where x and y variables can be calculated independently! From this, you can find “path” of particles Ex1) Given ax = f1(t) and ay = f2(t) Ex2) Given x = f1(t) and y = f2(t)

  33. y x Projectile motion • The most common case is when ax = 0 and ay = -g (approximation) • x and y direction can be calculated independently Note: a = const a=0 v vy vo vx vx (vo)y = v0 sin g vy v  (vo)x = v0 cos x-axis y-axis vx = (vo)x x = xo + (vo)xt ( in case of )

  34. y Determine the minimumhorizontal velocity u that a boy can throw a rock at A to just pass B. Note: rectilinear (a=const) (1) (2) (3) If we use eq. (3) in the y-direction : x g it can be applied in both x and y direction

  35. k : const Find x-,y-component of velocity and displacement as function of time , if the drag on the projectile results in an acceleration term as specified. Include the gravitational acceleration.

  36. Find R=R(q) y 2/95 Find q which maximizes R (in term of v-zero and a) x R=R(q) Find Rmax a

  37. H12-96 A boy throws 2 balls into the air with a speed v0 at the different angles {q1, q2} (q1 > q2). If he want the two ball collide in the mid air, what is the time delay between the 1st throw and 2nd throw. The first throw should be q1 or q2? q1 intersected point

  38. 2/84 Determine the maximum horizontal range R of the projectile and the corresponding launch angle q. this way? Yes! but why? h q=45 ? How do you throw the ball? h(q) = R(h) R Ceil’s height (5 m) is our limitation! R(q)

  39. 2/84 Determine the maximum horizontal range R of the projectile and the corresponding launch angle q. this way?

  40. H12/92 The man throws a ball with a speed v=15 m/s. Determine the angle q at which he should released the ball so that it strikes the wall at the highest point possible. The room has a ceiling height of 6m.

  41. t O n n n O t O t 2/5 Normal and Tangential Coordinates (n-t) Curves can be considered as many tangential circular arcs • takes positive tin the direction of increasing s Fixed point on curve Path: known • and positive ntoward the center of the curvature of the path s • the origin and the axes move (and rotate) along with the path of particle Forward velocity and forward acceleration make more sense to the driver The driver is only aware of forward direction (t) and lateral direction (n). Brake and acceleration force are often more convenient to describe relative to the car (t-direction). Turning (side) force also easier to describe relative to the car (n-direction)

  42. t O n n n O t O t Normal and Tangential Coordinates (n-t) generally, not total a Rectilinear Similarity Path: known v,a s s s Consider: scalar variables (s) along the path (t direction) The reason why we define this coordinate similar to rectilinear motion 0 why? s measured along the path

  43. y Velocity ( ): path x C  ds Velocity Small curves can be considered as circular arcs ** The velocity is always tangent to the path ** Path Speed: A d =  (d) A Fixed point on curve

  44. t n Acceleration Path t C (similarly)  db d A n ds = (d) A d d

  45. Using x-y coordinate Alternative Proof of Path y t C   n A O x

  46. n t d Understanding the equation Path C t  A d n ds = (d) A at comes from changes in the magnitude of an comes from changes in the direction of

  47. y path x What to remember by definition of n-t axis Rectilinear Similarity generally, not totala v,a s s

  48. y path x Proof y r db b+db ds dy b dx x

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