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MAE 242: DYNAMICS

MAE 242: DYNAMICS. Class 3. Dr. Samir N. Shoukry Professor, MAE & CEE. assistants K. Mc Bride Dhananjy Rao ony Oomen Praveen. Quiz 2 (5 minutes).

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MAE 242: DYNAMICS

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  1. MAE 242: DYNAMICS Class 3 Dr. Samir N. ShoukryProfessor, MAE & CEE assistants K. Mc Bride Dhananjy Raoony Oomen Praveen

  2. Quiz 2 (5 minutes) A car accelerates according to the relation a=0.02s m/s2. Determine its velocity when s=100 m if s=v=0 when t=0.

  3. Morgantown M M r r New York M/W M/W N N North North North North r r N/W N/W East East East East West West West West W W Point of reference that may be placed Point of reference that may be placed Point of reference that may be placed Point of reference that may be placed anywhere anywhere anywhere anywhere … … … … , say Charleston, WV , say Charleston, WV , say Charleston, WV , say Charleston, WV South South South South How Position May Be Described? Position is defined relative to a point on which we set a coordinate system

  4. rM/W Path Path rN/W rN/M Position Position Vector Vector New York New York Path Path N N rM/N W W E E New York S S Position Position Vector Vector OR Position is defined by a vector “r” that originates from a reference point Remember: A vector has a length and direction

  5. Rectangular Coordinates Position is defined by the distance from three perpendicular lines from an origin O. r = x i + y j r = x i + y j + z k

  6. r = x i + y j + z k vxi d d dx d dx di d i + x i (zk) (xi) = = = (yj) v= (xi) = + + dt dt dt dt dt dt dt dr dt v = vx i + vy j + vzk a = ax i + ay j + azk Rectangular Coordinates:Velocity & Acceleration

  7. y Example: Given , Calculate the magnitude and direction of the velocity and acceleration vectors. Calculate the magnitude of the position vector. 2

  8. Motion of A Projectile

  9. Class 4 MAE 242: DYNAMICS Instructor Dr. Samir N. ShoukryProfessor, MAE & CEE assistants Dr. G. William, CEE Mr. K. Mc Bride, MAE

  10. Quiz 3 (5 minutes) A particle moves according to the relation r=5t3i +2t2 j Calculate the magnitude and direction of its acceleration when t =0.5 second.

  11. Example: A particle accelerates according to the relation: Given the initial conditions at t=0: x, y, z =0 and vx=vy=1 & vz=0, calculate the particle velocity & position.

  12. Motion of A Projectile

  13. A skier leaves the ramp A at angle with the horizontal. He strikes the ground at point B, Determine his initial speed vAand the time of flight from A to B. -64

  14. y x

  15. 5 m y

  16. The boy at A attempts to throw a ball over the roof of a barn such that it is launched at an angle of 40 degrees. Determine the speed vA at which he must throw the ball so that it reaches its maximum height at C. Also, find the distance d where the boy must stand so that he can make the throw. Given: Y0 = 1m, At max. height, vY = 0, Y = 8m and v0x = v0cos400; v0y = v0sin400 (1) X = v0xt; (2) vx = v0x (constant velocity in x-direction); (3) Y = Y0 ­1/2gt2 + v0Yt (4) vY = v0Y ­ gt (constant acceleration -g in y-direction) At the top of the trajectory vy = 0 and from (4) we get: v0y = gt Substituting the above into (3) we get 7 = ½ g t2 and solving the above to get the time to reach maximum elevation. t = 1.195 seconds Substituting for v0Y at the top of the trajectory, we get  v0sin40o = 9.81(1.195) which is solved for the initial velocity. v0 = 18.24 m/sec Substituting into (1) we get the total distance X. X = 18.24(cos40)(1.195) = 16.7 m Which gives d = X ­ 4 = 12.70m

  17. The ball is thrown from the tower with a velocity of 20 ft/s as shown. Determine the x and y coordinates to where the ball strikes the slope. Also, determine the speed at which the ball hits the ground. Initial Conditions: v0 = 20 ft/s    v0x = 20(3/5) = 12 ft/s    v0y = 20(4/5) = 16 ft/s theta = 26.60 , X0 = 0       Y0 = 80 ft Motion in the x-direction at a constant velocity.  (1) vx = v0x = 12 ft/s (2) X = vxt= 12t Motion in the y-direction at a constant acceleration a = -g. Integrating for velocity (3) vy = v0y - gt = 16 - 32.2(t) and the y-displacement is obtained by integrating the latter, which gives Y = -1/2gt2 + v0yt + Y0 or (4) Y = -16.1t2 +16t + 80 The x and y coordinates of the ball when it hits the ground can be written as. (5) x = 20 + Rcos26.6o ; (6) y = Rsin26.6o Equations (2), (4) (5) and (6) are four equations in four unknowns. They can be solved for t - the time when the ball hits the ground. Thus t = 2.7 sec. Substituting into (2) and (4) provides the x and y coordinates of the ball when it hits the ground. x = 32.4 ft; y = 6.56 ft To solve for the speed, we substitute t = 2.7s into equation (3) which gives. vx = 12 ft/s and vy = 70.9 ft/s The total speed is found from v = (vx2 + vy2)1/2 ; v = 71.9ft/s

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