Dynamics Overview. Kinematics the study of a body s motion independent of the forces on the body.

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1 Overview 8-1 Dynamics the study of moving objects. Kinematics the study of a body s motion independent of the forces on the body. Kinetics the study of motion and the forces that cause motion.

2 Kinematics Rectangular Coordinates 8-2

3 Kinematics Polar Coordinates 8-3a1

4 Kinematics Polar Coordinates 8-3a2

5 Kinematics Circular Motion 8-4a

6 Kinematics Circular Motion 8-4b1 Angular velocity = Angular acceleration = Tangential acceleration = Normal acceleration =

7 Kinematics Circular Motion 8-4b2 Example (FEIM): A turntable starts from rest and accelerates uniformly at 1.5 rad/s 2. How many revolutions does it take for the rotational frequency to reach rpm? $ " = 2#f = 2# rad ' $ & ) rev ' $ & ) 1 min ' & ) = 3.49 rad % rev( % min( % 60 s ( s " = t f # $dt = 0 t f # %tdt = %t 2 f 0 2 t = " # " = #2 2$ ( ) 2 ( )( 1.5 rad s ) 2 rad 3.49 s = 2 = 4.06 rad n = 4.06 rad 2" rad rev = 0.65 revolution

8 Kinematics Projectile Motion 8-5a1 Constant acceleration formulas: s = s 0 + v 0 t at 2 v = v 0 + a 0 t ( ) v 2 = v a 0 s " s 0

9 Kinematics Projectile Motion 8-5a2 Example 1 (FEIM): A projectile is launched at 180 m/s at a 30 incline. The launch point is 150 m above the impact plane. Find the maximum height, flight time, and range.

10 Kinematics Projectile Motion 8-5a3 h max " = v 0 sin# + at = 0 % 180 m ( ' * 0.5 sin# t max = $v 0 a = $ & s ) % $9.8 m ( ' * & s ) ( ) = 9.18 s h = h 0 + v 0 t sin# + 1 at 2 2 % h max = 150 m m ( ' *(9.18 s)(0.5) $ 1 & s 2 2 ) h impact = 0 = 150 m + 90t $ 4.9t 2 By the quadratic equation, t impact cos30 = R impact = R 0 + v 0 t impact cos# % ( )' 9.8 m s 2 & = 19.9 s % = m ( ' *(19.9 s)(0.866) = 3100 m & s ) ( *(9.18 s) 2 = 563 m )

11 Kinematics Projectile Motion 8-5b1 Example 2 (FEIM): A bomber flies horizontally at 275 km/h and an altitude of 3000 m. At what viewing angle from the bomber to the target should the bomb be dropped?

12 Kinematics Projectile Motion 8-5b2 h impact = 1 2 at 2 = 1 2 2h impact # ( )%"9.8 m s 2 $ & ( t 2 = "3000 m ' t = = 24.7 s g # 275 km &# % $ hr ' ( 1000 m &# 1 hr & % $ km (% ( = m/s ' $ 3600 s' # R impact = R 0 + v 0 t = m & % ( 24.7 s $ s ' ) = arctan R impact h impact ( ) = 1887 m = arctan 1887 m 3000 m = 32.2

13 Kinetics Newton s 2nd Law of Motion 8-6a For a constant mass, One-dimension motion

14 Kinetics Newton s 2nd Law of Motion 8-6b1 Example (FERM prob. 6, p. 15-5):

15 Kinetics Newton s 2nd Law of Motion 8-6b2 Example (FERM prob. 6, p. 15-5):

16 Kinetics Impulse and Momentum 8-7a Impulse (constant mass in one dimension) Momentum Impulse-Momentum Principle

17 Kinetics Impulse and Momentum 8-7b1 Example 1 (FEIM): A kg marble attains a velocity of 76 m/s in a slingshot. Contact with the slingshot is 1/25 of a second. What is the average force on the marble during the launch? F ave = m"v "t = # (0.046 kg) % 76 m $ s 0.04 s & ( ' = 87.4 N

18 Kinetics Impulse and Momentum 8-7b2 Example 2 (FEIM): A 2000 kg cannon fires a 10 kg projectile horizontally at 600 m/s. It takes s for the projectile to pass through the barrel. What is the recoil velocity if the cannon is not restrained? What average force must be exerted on the cannon to keep it from moving? m projectile "v projectile = m cannon "v cannon (10 kg)(600 m s ) = (2000 kg)(v cannon) v cannon = 3 m s = initial recoil velocity F = m"v "t = (10 kg)(600 m s ) s = 8.57 #10 5 N

19 Work & Energy 8-8a Work Potential Energy Gravity Kinetic Energy of a Mass W = KE 2 " KE 1 = 1 m(v " v 12 ) W = PE 2 " PE 1 = mg(h 2 " h 1 ) Spring (linear) F s = kx where the spring is compressed a distance x Kinetic Energy of a Rotating Body W = PE 2 " PE 1 = 1 k(x " x 12 ) W = KE 2 " KE 1 = 1 I(# " # 12 )

20 Work & Energy 8-8b Conservation of Energy For a closed system (no external work), the change in potential energy equals the change in kinetic energy. PE 1 " PE 2 = KE 2 " KE 1 PE 1 +KE 1 = PE 2 +KE 2 For a system with external work, W equals ΔPE + ΔKE. W 1"2 = (PE 1 # PE 2 ) + (KE 2 # KE 1 ) PE 1 +KE 1 +W 1"2 = PE 2 +KE 2

21 Work & Energy Impacts: Momentum is always conserved. Elastic Impacts: Kinetic energy is conserved. Example 1 (FEIM): Two identical balls collide along their centerlines in an elastic collision. The initial velocity of ball 1 is 0.85 m/s. The initial velocity of ball 2 is 0.53 m/s. What is the relative velocity of each ball after the collision? (A) 0.53 m/s and 0.85 m/s (B) 0.72 m/s and 1.2 m/s (C) 5.1 m/s and 1.2 m/s (D) 0.98 m/s and 1.8 m/s 8-8c m 1 v 1 + m 2 v 2 = m 1 v " 1 + m 2 v " 2 m 1 = m 2 sov 1 + v 2 = v " 1 + v " 2 = 0.85 # 0.53 = m v m v = 1 m v " v 12 + v 22 = v " 1 + v " m v " Solving two equations and two unkowns: v " 1 = #0.53 m/s v " 2 = 0.85 m/s Therefore, (A) is correct.

22 Work & Energy 8-8d Example 2 (FEIM): Ball A of 200 kg is traveling at 16.7 m/s. It strikes stationary ball B of 200 kg along the centerline. What is the velocity of ball A after the collision? Assume the collision is elastic. (A) 16.7 m/s (B) 8.35 m/s (C) 0 (D) 8.35 m/s m A = m B sov A + v B = v " A + v " B = 16.7 m/s 2 2 v A2 + v B2 = v " A + v " B There are two possible solutions for these equations. v " A = 0, v " B = 16.7 m/s or v " A = 16.7 m/s, v " B = 0 Since there must be a change in the collision, ball A s velocity must be 0. Therefore, (C) is correct.

23 Work & Energy Inelastic Impacts: Kinetic energy does not have to be conserved if some energy is converted to another form. v " 1 # v " 2 = #e(v 1 # v 2 ) where e = coefficient of restitution v " 1 = m v (1+ e) + (m # em )v m 1 + m 2 v " 2 = m v (1+ e) + (em # m )v m 1 + m 2 8-8e Example 1 (FEIM): A ball is dropped from an initial height h o. If the coefficient of restitution is 0.90, how high will the ball rebound? (A) 0.45h o (B) 0.81h o (C) 0.85h o (D) 0.9h o mgh = 1 2 mv2 v = 2gh o v " = #ev = #e 2gh o = 2g h " h " = e 2 h o = (0.9) 2 h o = 0.81h o Since there must be a change in the collision, ball A s velocity must be 0. Therefore, (B) is correct.

24 Work & Energy 8-8f Example 2 (FEIM): Two masses collide in a perfectly inelastic collision. What is the velocity of the combined mass after collision? m 1 = 4m 2 v 1 = 10 m/s v 2 = "20 m/s (A) 0 (B) 4 m/s (C) 5m/s (D) 10 m/s Perfectly inelastic means the masses collide and stick together. m 1 v 1 + m 2 v 2 = m 3 v 3 m 3 = m 1 + m 2 = 5m 2 " 4m 2 10 m % " $ ' + m 2 (20 m % $ ' = 5m 2 v 3 # s & # s & " 5m 2 v 3 = 40 m % $ # s & ' m ( " 20 m % $ 2 s ' m 2 # & v 3 = 4 m/s Therefore, (B) is correct.

25 Kinetics 8-9a Friction Example (FEIM): A snowmobile tows a sled with a weight of 3000 N. It accelerates up a 15 slope at 0.9 m/s 2. The coefficient of friction between the sled and the snow is 0.1. What is the tension in the tow rope? F slope = F rope " (F friction + F gravity ) = ma slope F rope = F friction + F gravity + ma slope = mg sin15 + mgµ cos15 + ma slope = (3000)(0.2588) + (3000 N)(0.1)(0.9659) " % $ N '" $ $ 9.8 m ' 0.9 m % $ ' '# s 2 & # s 2 & = 1342 N

26 Kinetics 8-9b1 Plane Motion of a Rigid Body Similar equations can be written for the y-direction or any other coordinate direction. Example (FEIM): A 2500 kg truck skids with a deceleration of 5 m/s 2. What is the coefficient of sliding friction? What are the frictional forces and normal reactions (per axle) at the tires?

27 Kinetics 8-9b2 The force of deceleration is equal to the friction force. " F deceleration = (2500 kg) 5 m % $ ' = F s 2 friction = µmg # & " = µ(2500 kg) 9.8 m % $ ' # s 2 & " µ = 5 m %" $ # s 2 & ' 9.8 m % $ s ' = 0.51 # 2 & The moment about the center of gravity, M A, must be equal to zero. " M A = (4.5 m)n B # (2 m)mg # (1 m)(f deceleration ) = 0 # = (4.5 m)n B " (2 m)(2500 kg) % 9.8 m & $ s 2 ' (" (2500 kg) # % 5 m & $ s ( = 0 2 ' N B = 13,667 N # N A = mg " N B = (2500 kg) % 9.8 m & ("13667 N = N $ s 2 '

28 Rotation 8-10a1 Rotation about Fixed Axis

29 Rotation Example (FEIM): A mass m is attached to a rope wound around a cylinder of mass m C and radius r. What is the acceleration of the falling mass? What is the rope tension? 8-10a2 There are three simultaneous equations for the movement of the mass, the cylinder, and the relationship between the two: mg " F = ma Fr = I# #r = a

30 Rotation 8-10a3 Rearranging, Fr = I a r F = m C r 2 a 2r 2 = m C 2 a mg " m C 2 a = ma a = g m m + m C 2 Solving for F, F = m C 2 a = g m C m " 2 m + m C $ # 2 % ' & = mm C 2m + m C g

31 Rotation 8-10b Centripetal Force The force required to keep a body rotating about an axis. (r is the distance from the center of mass to the center of rotation.) Centrifugal Force The reaction to centripetal force. The centrifugal force, like any inertia force, should not be used in free-body diagrams. Example (FEIM): A 2000 kg car travels 65 km/hr around a curve of radius 60 m. What is the centripetal force? 65 km/hr = m/s F c = m v2 r = " (2000 kg) m % $ ' # s & 60 m 2 = N

32 Rotation 8-10c Banking Curves Example (FEIM): A 2000 kg car travels at 64 km/hr around a banked curve with a radius of 150 m. What should the angle between the roadway and the horizontal be so tire friction is not needed to prevent sliding? " 64 km %" $ # hr & ' 1000 m %" 1hr % $ # km ' $ ' = 17.8 m &# 3600 s& s " " 17.8 m % % " % $ $ ' ' s ( = arctan$ v2 # & ' = arctan$ ' # gr & " $ 9.8 m = 12.1 $ % ' $ '(150 m) ## s ' & &

33 Rotation Free Vibration Spring and Mass: 8-10d Natural Frequency: Solution: For initial conditions x(0) = x 0 and x'(0) = v 0, For initial conditions x(0) = x 0 and x'(0) = 0, Torsional Free Vibration: Natural frequency: " n = k t I Solution:

34 Rotation 8-10e1 Example (FEIM): A 54 kg mass is supported by three springs, as shown. The starting position is 5.0 cm down from the equilibrium position. No external forces act on the mass after it is released. What are the maximum velocity and acceleration?

35 Rotation 8-10e2 From the solution to the differential equation, # x(t) = x 0 cos " n t + v & 0 % ( sin" n t $ ' " n v(t) = x )(t) = *x 0 " n sin" n t + v 0 cos " n t But, v 0 = 0; so v(t) = "x 0 # n sin# n t x 0 = 0.05 m k = k 1 + k 2 + k 3 = 1750 N m N m N m = 7875 N m N 7875 k " n = m = m = rad/s 54 kg % v = x # = ($0.05 m) rad ( ' * sin12.08t & s ) = "0.604 m/s sin12.08t

36 Rotation 8-10e3 The maximum velocity is when sin 12.08t = 1. Because the motion is oscillatory, the maximum velocity occurs in both directions. v max = ±0.604 m/s a = v " = (#0.05 m) rad 2 $ ' & ) cos12.08t % s ( a max = ±7.30 m/s 2

37 Review 8-11a1 Example: A yoyo (mass m, inertia I about center of gravity) is placed on a horizontal surface with a coefficient of friction µ. The string, wrapped underneath, is pulled with a force F. Determine if the yoyo will slip on the floor and which direction it will rotate as a function of F. The two equations for the movement of the center of gravity and the rotation are F + R = ma Fr i + Rr o = I"

38 Review 8-11a2 There are two possibilities for the third equation: slip or no slip. No slip: R < µmg " a = #r o $ Solving these 3 equations with 3 unknowns (α, a, and R) leads to " = #F r o # r i I + mr o 2 $" < 0 ( ) a = Fr r " r o o i #a > 0 2 I + mr o R = "F I + mr 2 i 2 I + mr o The yoyo is moving to the right while wrapping itself on the string. The solution for R gives the condition for no slip. F < µmg I + mr o 2 I + mr i 2

39 Review 8-11a3 Slip: R = µmg Solving these 3 equations lead to a = F " µmg m = Fr i " µmgr o I The yoyo always moves to the right, but can rotate forward or backard depending on the value for F.