Dynamics 8-1 Overview 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. Professional Publications, Inc. FERC
Dynamics 8-2 Kinematics—Rectangular Coordinates Professional Publications, Inc. FERC
Dynamics 8-3a1 Kinematics—Polar Coordinates Professional Publications, Inc. FERC
Dynamics 8-3a2 Kinematics—Polar Coordinates Professional Publications, Inc. FERC
Dynamics 8-4a Kinematics—Circular Motion Professional Publications, Inc. FERC
Dynamics 8-4b1 Kinematics—Circular Motion Angular velocity = Angular acceleration = Tangential acceleration = Normal acceleration = Professional Publications, Inc. FERC
Dynamics 8-4b2 Kinematics—Circular Motion 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 33.33 rpm? � � � � � � � = 2 � f = 2 � rad � 33.33 rev � 1 min � = 3.49rad � � � rev min 60 s s � � � � � � � � tdt = � t f 2 t t � � dt = f f � = 2 0 0 t = � � 2 ( 3.49 rad ) 2 � = � s = 4.06 rad 2 � = ( ) 1.5 rad 2 ( ) s 2 n = 4.06 rad = 0.65 revolution 2 � rad rev Professional Publications, Inc. FERC
Dynamics 8-5a1 Kinematics—Projectile Motion Constant acceleration formulas: s = s 0 + v 0 t + 1 2 at 2 v = v 0 + a 0 t 2 = v 0 2 + 2 a 0 s � s 0 v ( ) Professional Publications, Inc. FERC
Dynamics 8-5a2 Kinematics—Projectile Motion 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. Professional Publications, Inc. FERC
Dynamics 8-5a3 Kinematics—Projectile Motion h max = v 0 sin � + at = 0 � � 180m � ( ) � 0.5 � s sin � � � t max = � v 0 = 9.18 s = � a � � � 9.8m � � s � � h = h 0 + v 0 t sin � + 1 2 at 2 � � � � h max = 150 m + 180 m ( ) 9.8 m 2 = 563 m (9.18 s)(0.5) � (9.18 s) 1 � � � � s 2 s 2 2 � � � � h impact = 0 = 150 m + 90 t � 4.9 t 2 By the quadratic equation, t impact = 19.9 s cos30 ° = 0.866 R impact = R 0 + v 0 t impact cos � = 0 + 180m � � (19.9 s)(0.866) = 3100 m � � s � � Professional Publications, Inc. FERC
Dynamics 8-5b1 Kinematics—Projectile Motion 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? Professional Publications, Inc. FERC
Dynamics 8-5b2 Kinematics—Projectile Motion � � ( ) � 9.8 m 2 = 2 = � 3000 m h impact = 1 2 at t 1 � � 2 s 2 � � 2 h impact t = = 24.7 s g � � � � � � 275km � 1000 m � 1 hr � = 76.39 m/s � � � hr km 3600 s � � � � � � R impact = R 0 + v 0 t = 0 + 76.39m � � ( ) = 1887 m � 24.7 s � s � � � = arctan R impact = arctan1887 m 3000 m = 32.2 ° h impact Professional Publications, Inc. FERC
Dynamics 8-6a Kinetics—Newton’s 2nd Law of Motion For a constant mass, One-dimension motion Professional Publications, Inc. FERC
Dynamics 8-6b1 Kinetics—Newton’s 2nd Law of Motion Example (FERM prob. 6, p. 15-5): Professional Publications, Inc. FERC
Dynamics 8-6b2 Kinetics—Newton’s 2nd Law of Motion Example (FERM prob. 6, p. 15-5): Professional Publications, Inc. FERC
Dynamics 8-7a Kinetics—Impulse and Momentum Impulse (constant mass in one dimension) Momentum Impulse-Momentum Principle Professional Publications, Inc. FERC
Dynamics 8-7b1 Kinetics—Impulse and Momentum Example 1 (FEIM): A 0.046 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? (0.046 kg) 76m � � � � s ave = m � v � � F = 87.4 N = � t 0.04 s Professional Publications, Inc. FERC
Dynamics 8-7b2 Kinetics—Impulse and Momentum Example 2 (FEIM): A 2000 kg cannon fires a 10 kg projectile horizontally at 600 m/s. It takes 0.007 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)(600m s ) = (2000 kg)(v cannon ) v cannon = 3m s = initial recoil velocity (10 kg)(600m s ) F = m � v = 8.57 � 10 5 N = � t 0.007 s Professional Publications, Inc. FERC
Dynamics 8-8a Work & Energy Work Potential Energy • Gravity W = PE 2 � PE 1 = mg ( h 2 � h 1 ) Kinetic Energy of a Mass • Spring (linear) F s = kx where the spring is 2 � v 1 W = KE 2 � KE 1 = 1 2 m (v 2 2 ) compressed a distance x Kinetic Energy of a Rotating Body 2 � x 1 W = PE 2 � PE 1 = 1 2 k ( x 2 2 ) 2 � � 1 W = KE 2 � KE 1 = 1 2 I ( � 2 2 ) Professional Publications, Inc. FERC
Dynamics 8-8b Work & Energy 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 Δ PE + Δ KE. W 1 � 2 = (PE 1 � PE 2 ) + (KE 2 � KE 1 ) PE 1 + KE 1 + W 1 � 2 = PE 2 + KE 2 Professional Publications, Inc. FERC
Dynamics 8-8c Work & Energy m 1 v 1 + m 2 v 2 = m 1 � v 1 + m 2 � v Impacts: Momentum is always 2 conserved. m 1 = m 2 so v 1 + v 2 = � v v 2 = 0.85 � 0.53 = 0.32 1 + � Elastic Impacts: Kinetic energy is 2 + 1 2 + 1 2 = 1 2 2 m v 1 2 m v 2 2 m � v 2 m � v 1 conserved. 1 2 2 + � 2 + v 2 2 = � 2 v 1 v v 1 2 Solving two equations and two unkowns: Example 1 (FEIM): v 1 = � 0.53 m/s � Two identical balls collide along v 2 = 0.85 m/s � their centerlines in an elastic collision. The initial velocity of Therefore, (A) is correct. 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 Professional Publications, Inc. FERC
Dynamics 8-8d Work & Energy 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 so v A + v B = � v v B = 16.7 m/s A + � 2 + � 2 + v B 2 = � 2 v A v v A 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. Professional Publications, Inc. FERC
Dynamics 8-8e Work & Energy Inelastic Impacts: Example 1 (FEIM): Kinetic energy does not have to be A ball is dropped from an initial height conserved if some energy is converted h o . If the coefficient of restitution is to another form. 0.90, how high will the ball rebound? (A) 0.45 h o v v 2 = � e (v 1 � v 2 ) (B) 0.81 h o 1 � � � (C) 0.85 h o where e = coefficient of restitution (D) 0.9 h o 1 = m 2 v 2 (1 + e ) + ( m 1 � em 2 )v 1 v � mgh = 1 2 m v 2 m 1 + m 2 2 = m 1 v 1 (1 + e ) + ( em 1 � m 2 )v 2 v = 2 gh o v � m 1 + m 2 v = � e v = � e 2 gh o = 2 g � h � h = e 2 h o = (0.9) 2 h o = 0.81 h o � Since there must be a change in the collision, ball A’s velocity must be 0. Therefore, (B) is correct. Professional Publications, Inc. FERC
Dynamics 8-8f Work & Energy Example 2 (FEIM): Two masses collide in a perfectly inelastic collision. What is the velocity of the combined mass after collision? m 1 = 4 m 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 = 5 m 2 4 m 2 10m � � � + m 2 � 20m � � � = 5 m 2 v 3 � � s s � � � � � � � � 5 m 2 v 3 = 40m m 2 � 20m m 2 � � � � s s � � � � v 3 = 4 m/s Therefore, (B) is correct. Professional Publications, Inc. FERC
Dynamics 8-9a Kinetics 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 °+ mg µ cos15 °+ ma slope = (3000)(0.2588) + (3000 N)(0.1 )(0.9659) � � � � + 3000 N � 0.9 m � � � � � 9.8 m s 2 � � � � s 2 � � = 1342 N Professional Publications, Inc. FERC
Dynamics 8-9b1 Kinetics 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? Professional Publications, Inc. FERC
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