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FE Review

FE Review. Dynamics G. Mauer UNLV Mechanical Engineering. Point Mass Dynamics. X-Y Coordinates. 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/s 2. Use g = 10 m/s 2.

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FE Review

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  1. FE Review Dynamics G. Mauer UNLVMechanical Engineering

  2. Point Mass Dynamics X-Y Coordinates

  3. 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

  4. 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

  5. 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

  6. 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

  7. 12.7 Normal and Tangential Coordinates ut : unit tangent to the path un : unit normal to the path

  8. Normal and Tangential Coordinates Velocity Page 53

  9. Normal and Tangential Coordinates

  10. 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

  11. 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

  12. Polar coordinates

  13. Polar coordinates

  14. Polar coordinates

  15. 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

  16. 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

  17. 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

  18. 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

  19. Example cont’d: Problem 2.198 Sailboat tacking against Northern Wind 2. Vector equation (1 scalar eqn. each in i- and j-direction) 500 150 i

  20. 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

  21. 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

  22. 2.9 Constrained Motion vA is given as shown. Find vB Approach: Use rel. Velocity: vB = vA +vB/A (transl. + rot.)

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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.

  28. 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

  29. 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 )

  30. 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

  31. 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

  32. Mass m rests on the 30 deg. Incline as shown. Step 1: Free-Body Analysis. Step 2: Apply Newton’s Law in each Direction: M*g*sinq*i N -M*g*cosq*j M*g

  33. Friction F = mk*N: Another horizontal reaction is added in negative x-direction. M*g*sinq*i mk*N N -M*g*cosq*j M*g

  34. Problem 3.27 in Book: Find accel of Mass A Start with: • Newton’s Law for A. • Newton’s Law for A and B • Free-Body analysis of A and B • Free-Body analysis of A

  35. Problem 3.27 in Book: Find accel of Mass A Start with: • Newton’s Law for A. • Newton’s Law for A and B • Free-Body analysis of A and B • Free-Body analysis of A

  36. Problem 3.27 in Book cont’d Newton applied to mass B gives: • SFu = 2T = mB*aB • SFu = -2T + mB*g = 0 • SFu = mB*g-2T = mB*aB • (D) SFu = 2T- mB*g-2T = 0

  37. Problem 3.27 in Book cont’d Newton applied to mass B gives: • SFu = 2T = mB*aB • SFu = -2T + mB*g = 0 • SFu = mB*g-2T = mB*aB • (D) SFu = 2T- mB*g-2T = 0

  38. Problem 3.27 in Book cont’d Newton applied to mass A gives: • SFx = T +F= mA*ax ; SFy = N - mA*g*cos(30o) = 0 • SFx = T-F= mA*axSFy = N- mA*g*cos(30o) = mA*ay • SFx = T = mA*ax ; SFy = N - mA*g*cos(30o) =0 • (D) SFx = T-F = mA*ax ; SFy = N-mA*g*cos(30o) =0

  39. Problem 3.27 in Book cont’d Newton applied to mass A gives: • SFx = T +F= mA*ax ; SFy = N - mA*g*cos(30o) = 0 • SFx = T-F= mA*axSFy = N- mA*g*cos(30o) = mA*ay • SFx = T = mA*ax ; SFy = N - mA*g*cos(30o) =0 • (D) SFx = T-F = mA*ax ; SFy = N-mA*g*cos(30o) =0

  40. Energy Methods

  41. Only Force components in direction of motion do WORK

  42. Work of Gravity

  43. Work of a Spring

  44. The work-energy relation: The relation between the work done on a particle by the forces which are applied on it and how its kinetic energy changes follows from Newton’s second law.

  45. A car is traveling at 20 m/s on a level road, when the brakes are suddenly applied and all four wheels lock. mk = 0.5. The total distance traveled to a full stop is (use Energy Method, g = 10 m/s2) • 40 m • 20 m • 80 m • 10 m • none of the above

  46. Collar A is compressing the spring after dropping vertically from A. Using the y-reference as shown, the work done by gravity (Wg) and the work done by the compression spring (Wspr) are (A) Wg <0, Wspr <0 (B) Wg >0, Wspr <0 (C) Wg <0, Wspr >0 (D) Wg >0, Wspr >0 y

  47. Conservative Forces A conservative force is one for which the work done is independent of the path taken Another way to state it: The work depends only on the initial and final positions,not on the route taken.

  48. Conservative Forces T1 + V1 = T2 + V2

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