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# Final Exam Review

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. Télécharger la présentation ## Final Exam Review

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1. Final Exam Review

2. Review Always work from first Principles!

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

4. Review 1. Free-Body

5. B_x mg B_y 1. Free-Body

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

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

8. mg A_x A_y N 1. Free-Body

9. 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

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

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

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

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

14. Point Mass Dynamics X-Y Coordinates

15. 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

16. 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

17. 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

18. 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

19. 12.7 Normal and Tangential Coordinates ut : unit tangent to the path un : unit normal to the path

20. Normal and Tangential Coordinates Velocity Page 53

21. Normal and Tangential Coordinates

22. 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

23. 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

24. Polar coordinates

25. Polar coordinates

26. Polar coordinates

27. 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

28. 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

29. 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

30. 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

31. 12.10 Relative (Constrained) Motion We Solve Graphically (Vector Addition) vA vB vB/A

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

33. 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

34. 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

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

36. 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

37. 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

38. 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

39. 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

40. 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

41. 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

42. 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.

43. 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

44. 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 )

45. 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

46. 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

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