SlideServe
Télécharger Obtenir Pro
Menu
Accueil Pro
Préférences
ApparenceLight Aide et commentaires
© SlideServe · v2026

Final Exam Review

thom avatar
thom
102 diapositives · 1.3K Views · Aug 20, 2014

Final Exam Review. Please Return Loan Clickers to the MEG office after Class! Today!. Review. Always work from first Principles!. Review. Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints. Review. 1. Free-Body. B_x. mg. B_y. 1. Free-Body. B_x.

#final exam review #final exam review ppt #powerpoint #presentation
Final Exam Review - slide 1
01 / 102Final Exam Review
1 / 17

Télécharger la présentation

Final Exam Review
Une image / un lien ci-dessous est fourni (tel quel) pour télécharger la présentation

Politique de téléchargement : le contenu du site vous est fourni TEL QUEL pour votre information et votre usage personnel ; il ne peut être vendu, concédé sous licence ni partagé sur d’autres sites sans l’accord de son auteur.Le contenu vous est fourni TEL QUEL, pour votre information et votre usage personnel uniquement. Téléchargez la présentation en cliquant sur ce lien. Lors du téléchargement, si vous ne parvenez pas à télécharger une présentation, l’éditeur a peut-être supprimé le fichier de son serveur.Pendant le téléchargement, si vous ne parvenez pas à obtenir une présentation, le fichier a peut-être été supprimé par l’éditeur.

F I N

Plus de contenu associé

Transcription de la présentation

  1. Final Exam Review

  2. Please Return Loan Clickers to the MEG office after Class!Today!

  3. Review Always work from first Principles!

  4. Review Always work from first Principles! Kinetics: Free-Body Analysis Newton’s Law Constraints

  5. Review 1. Free-Body

  6. B_x mg B_y 1. Free-Body

  7. B_x mg B_y 2. Newton Moments about B: -mg*L/2 = IB*a with IB = m*L2/3

  8. B_x mg B_y 3. Constraint aG = a*L/2 = -g*3/4

  9. mg A_x A_y N 1. Free-Body

  10. mg 2. Newton A_x A_y N Moments about Center of Cylinder:A_x From triangle at left: Ax*(R-h) –b*mg = 0 acart*(R-h) –b*g = 0

  11. mg 2. Newton A_x A_y N N = 0 at impending rolling, thus Ay = mg Ax = m*acart

  12. Kinematics (P. 16-126) CTR

  13. Kinematics (P. 16-126) 4r -2r*i + 2r*j CTR

  14. Feedback • Overall, when comparing traditional Homework formats with Mastering, I prefer • Paper submission of Homework • Electronic Submission

  15. Feedback • For me, the most useful benefit of Mastering is • Hints while developing the solution to a problem • Instant grading of results • Practice Exams

  16. Point Mass Dynamics X-Y Coordinates

  17. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters. • The travel time t to Point B is • t = 4 s • t = 1 s • t = 0.5 s • (D)t = 2 s • Use g = 10 m/s2 Use g = 10 m/s2

  18. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters. • The travel time t to Point B is • t = 4 s • t = 1 s • t = 0.5 s • (D)t = 2 s • Use g = 10 m/s2 Use g = 10 m/s2

  19. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s. • The start velocity v0 is • v0 = 40 m/s • v0 = 20 m/s • v0 = 10 m/s • (D) v0 = 5 m/s Use g = 10 m/s2

  20. A ball is thrown horizontally from A and passes through B(d=20,h = -20) meters at time t = 2s. • The start velocity v0 is • v0 = 40 m/s • v0 = 20 m/s • v0 = 10 m/s • (D) v0 = 5 m/s Use g = 10 m/s2

  21. 12.7 Normal and Tangential Coordinates ut : unit tangent to the path un : unit normal to the path

  22. Normal and Tangential Coordinates Velocity Page 53

  23. Normal and Tangential Coordinates

  24. Fundamental Problem 12.27 The boat is traveling along the circular path with r = 40m and a speed of v  = 0.5*t2  , where t is in seconds. At t = 4s, the normal acceleration is: • (A) constant • (B) 1 m/s2 • (C) 2 m/s2 • (D) not enough information • (E) 4 m/s2

  25. Fundamental Problem 12.27 The boat is traveling along the circular path with r = 40m and a speed of v  = 0.5*t2  , where t is in seconds. At t = 4s, the normal acceleration is: • (A) constant • (B) 1 m/s2 • (C) 2 m/s2 • (D) not enough information • (E) 4 m/s2

  26. Polar coordinates

  27. Polar coordinates

  28. Polar coordinates

  29. Polar Coordinates • Point P moves on a counterclockwise circular path, with r =1m, q-dot(t) = 2 rad/s. The radial and tangential accelerations are: • (A) ar = 4m/s2 aq = 2 m/s2 • (B) ar = -4m/s2 aq = -2 m/s2 • (C) ar = -4m/s2 aq = 0 m/s2 • (D) ar = 0 m/s2 aq = 0 m/s2

  30. Polar Coordinates • Point P moves on a counterclockwise circular path, with r =1m, q-dot(t) = 2 rad/s. The radial and tangential accelerations are: • (A) ar = 4m/s2 aq = 2 m/s2 • (B) ar = -4m/s2 aq = -2 m/s2 • (C) ar = -4m/s2 aq = 0 m/s2 • (D) ar = 0 m/s2 aq = 0 m/s2

  31. Point B moves radially outward from center C, with r-dot =1m/s, q-dot(t) = 10 rad/s. At r=1m, the radial acceleration is: • (A) ar = 20 m/s2 • (B) ar = -20 m/s2 • (C) ar = 100 m/s2 • (D) ar = -100 m/s2

  32. Point B moves radially outward from center C, with r-dot =1m/s, q-dot(t) = 10 rad/s. At r=1m, the radial acceleration is: • (A) ar = 20 m/s2 • (B) ar = -20 m/s2 • (C) ar = 100 m/s2 • (D) ar = -100 m/s2

  33. 12.10 Relative (Constrained) Motion We Solve Graphically (Vector Addition) vA vB vB/A

  34. Example : Sailboat tacking against Northern Wind 2. Vector equation (1 scalar eqn. each in i- and j-direction) 500 150 i

  35. Given: • r(t) = 2+2*sin(q(t)), q_dot= constant • The radial velocity is • 2+2*cos(q(t ))*q-dot, • -2*cos(q(t))*q-dot • 2*cos(q(t))*q-dot • 2*cos(q(t)) • 2*q +2*cos(q(t ))*q-dot

  36. Given: • r(t) = 2+2*sin(q(t)), q_dot= constant • The radial velocity is • 2+2*cos(q(t ))*q-dot, • -2*cos(q(t))*q-dot • 2*cos(q(t))*q-dot • 2*cos(q(t)) • 2*q+2*cos(q(t ))*q-dot

  37. Constrained Motion vA is given as shown. Find vB Approach: Use rel. Velocity: vB = vA +vB/A (transl. + rot.)

  38. The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is • 6 m/s • 40 rad/s • -40 rad/s • 4 rad/s • none of the above

  39. The conveyor belt is moving to the left at v = 6 m/s. The angular velocity of the drum (Radius = 150 mm) is • 6 m/s • 40 rad/s • -40 rad/s • 4 rad/s • none of the above

  40. Omit all constants! • The rope length between points A and B is: • (A) xA – xB + xc • (B) xB – xA + 4xc • (C) xA – xB + 4xc • (D) xA + xB + 4xc

  41. Omit all constants! • The rope length between points A and B is: • (A) xA – xB + xc • (B) xB – xA + 4xc • (C) xA – xB + 4xc • (D) xA + xB + 4xc

  42. Given: v0 = const. • The vertical velocity component of point A (in y-direction) is • vA,y = v0*tan(q) • vA,y = v0*cot(q) • vA,y = v0*cos(q) • vA,y = 2*v0 q q

  43. Vy/vx =cot q q • Given: v0 = const. • The velocity of point A in vertical y-direction is • vA,y = v0*tan(q) • vA,y = v0*cot(q) • vA,y = v0*cos(q) • vA,y = 2*v0 • vA,y = v0/cos(q) q q

  44. NEWTON'S LAW OF INERTIA A body, not acted on by any force, remains in uniform motion. NEWTON'S LAW OF MOTION Moving an object with twice the mass will require twice the force. Force is proportional to the mass of an object and to the acceleration (the change in velocity). F=ma.

  45. Dynamics M1: up as positive: Fnet = T - m1*g = m1 a1 M2: down as positive. Fnet = F = m2*g - T = m2 a2 3. Constraint equation: a1 = a2 = a

  46. Equations From previous: T - m1*g = m1 a  T = m1 g + m1 a Previous for Mass 2: m2*g - T = m2 a Insert above expr. for T m2 g - ( m1 g + m1 a ) = m2 a ( m2 - m1 ) g = ( m1 + m2 ) a ( m1 + m2 ) a = ( m2 - m1 ) g a = ( m2 - m1 ) g / ( m1 + m2 )

  47. Rules 1. Free-Body Analysis, one for each mass 2. Constraint equation(s): Define connections. You should have as many equations as Unknowns. COUNT! 3. Algebra: Solve system of equations for all unknowns

  48. Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Best approach: use coordinates tangential and normal to the path of motion as shown. M*g*sinq*i -M*g*cosq*j M*g