Table of Contents g. # 1 1/11/16 Momentum & Impulse (Bozemanscience Videos) 2 1/13/16 Conservation of Momentum 3 1/19/16 Elastic and Inelastic Collisions 4 1/19/16 Lab 1 Momentum 5 1/26/16 Rotational Dynamics
Chapter 8 Rotational Kinematics Rolling motion involves both linear and rotational motion. If there is no slipping at the point of contact with the ground, the motion of the tire of an car is an example of rolling motion. While the car is moving with linear speed v covering a distance d, a point on the outer edge of the tire moves the same distance along the circular path s = θ r = d. Thus, the linear speed v = d t of the car is the same as the tangential speed of a point on the outer edge of the tire: v T = s t = θ r t = ( θ t) r = ω r r Linear velocity, v (a) B s B d = s
Chapter 8 Rotational Kinematics Vector Nature of ngular Variables Like linear velocity and acceleration, also angular velocity (ω) and angular acceleration (α) are vector quantities. They both have magnitude and direction ω = Δθ Δt = θ f θ i Δt radians second ω Right hand α = Δω Δt = ω f ω i Δt radians second 2 Right hand ω 8.16
Warm Up (2/01/16) Rotational Kinematics person lowers a bucket into a well by turning roblem the hand 40 crank, as the drawing illustrates. The crank handle moves with a constant tangential speed of 1.20 m/s on its circular path. The rope holding the bucket unwinds without slipping on the barrel of the crank. ind the linear speed with which the bucket moves down the well. Given: v hand crank = 1.20 m/s v hand = ω r hand crank crank ω = v hand crank r hand crank v = r ω = ( 0.100 m 2 ) (6 rad/s) = 0.3 m/s a = r α 0.100 m diameter 0.400 m diameter ω = 1.20 m/s = 6 rad/s 0.200 m = u 1
Chapter 9 Rotational Dynamics What causes an object to have an angular acceleration? Torque, τ = (Magnitude of the force) (Lever rm) = l [ N m]
Chapter 9 Rotational Dynamics What causes an object to have an angular acceleration? Torque, τ = (Magnitude of the force) (Lever rm) = l [ N m] inge (axis of rotation) B Door B ( a) ( b) B ( c) igure 9.2 With a force of a given
Chapter 9 Rotational Dynamics What causes an object to have an angular acceleration? Torque, τ = (Magnitude of the force) (Lever rm) = l [ N m] B Line of action Line of action Line of action xis of Rotation 90 B B Lever arm = 90 = 0, since line of action passes through axis
Chapter 9 Rotational Dynamics Determine the lever arm of the torque exerted by each force. Torque, τ = (Magnitude of the force) (Lever rm) = l [ N m] xis of rotation l 4 = 0 l 1 = 0 Door (overhead view) l 2 = 0 l 3 = 0 1 2 3 4
Chapter 9 Rotational Dynamics Determine the lever arm of the torque exerted by each force. Torque, τ = (Magnitude of the force) (Lever rm) = l [ N m] l = 0 Gy 2.30 m 4.00 m 50.0 50.0 l L = 0 l = 0 W W L G x -body diagram ( a) of the ladder
Chapter 9 Rotational Dynamics The tendon exerts a force of magnitude 790 N. Determine the torque (magnitude and direction) of this force about the ankle joint. τ = l = 790N [3.6 10 2 m Cos(55 )] = 15 N m! cos55 " = 3.6" 10! 2 m
Chapter 9 Rotational Dynamics If a rigid body is in equilibrium, it has zero translational acceleration and zero angular acceleration. net = ma; system: diving board ext 1 + 2 + mg = ma 1 + 2 mg = 0 0, equilibrium Bolt 1 =? 1.4 m 2 =? 3.9 m ulcrum +y g = mg = -530 N + CONCET T GLNCE External orces 1. Gravitational orce (ection 4.7) 2. Normal orce (ection 4.8) 3. rictional orces (ection 4.9) 4. Tension orce (ection 4.10) um of External orces Is Zero: Σx = 0 Σy = 0 (4.9a) (4.9b) um of External Torques Is Zero: Στ = 0 (9.2) Rigid Bodies in Equilibrium
Chapter 9 Rotational Dynamics If a rigid body is in equilibrium, it has zero translational acceleration and zero angular acceleration. net = ma; system: diving board ext 1 + 2 + mg = ma 1 + 2 mg = 0 τ 1 + τ 2 + τ = 0 g 0, equilibrium Bolt 1 =? 1.4 m 2 2 = 1 mg =? 3.9 m ulcrum 2 = 946 N 530N 2 = 1,476 N τ 1 + 0 τ = 0 g τ 1 = τ g l 1 1 = l w mg 1 = l w mg l 1 = 2.5m (530N) 1.4 m +y g = mg = -530 N + = 946 N
Warm Up (02/02/16) Equilibrium n 8.00-m ladder of weight W = 355 N leans against a smooth vertical wall. The term smooth means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. firefighter, whose weight is W = 875 N, stands 6.30 m up from the bottom of the ladder. ssume that the ladder s weight acts at the ladder s center and neglect the hose s weight. ind the forces that the wall and the ground exert on the ladder. Gy 2.30 m net = ma; system: ladder, m L ext G + W L + W + = ma ( G x + ) ^ x + (G y W L W ) ^y = 0 ( G x + ) = 0 & (G y W L W ) = 0 4.00 m W W L 50.0 50.0 G +y x -body diagram ( a) of the ladder +
τ net = 0 ext Warm Up (02/02/16) Equilibrium τ G + τ W + τ W + τ = 0 L net = ma; system: ladder, m L ext G + W L + W + = ma 0 l 1 W l 2 W + l 3 = 0 L 4mCos(50 ) 355N 6.3mCos(50 )875N + 8.0m in(50 ) = 0 4456 N m + 8.0m in(50 ) = 0 = 4456 N m 8.0m in(50 ) = 727 N ( G x + ) ^ x + (G y W L W ) = 0 ( G x + ) = 0 & (G y W L W ) = 0 y ^ Gy +y 4.00 m + 50.0 50.0 W L G x -body diagram ( a) of the ladder 2.30 m W
Warm Up (02/03/16) Equilibrium woman whose weight is 530 N is poised at the right end of a diving board with a length of 3.90 m. The board has negligible weight and is bolted down at the left end, while being sup- ported 1.40 m away by a fulcrum. ind the forces 1 and 2 that the bolt and the fulcrum, respectively, exert on the board. net = ma; system: diving board ext 1 + 2 + mg = ma 1 + 2 mg = 0 0, equilibrium Bolt 1 =? 2 =? 3.9 m +y + 1.4 m ulcrum g = mg =
Warm Up (02/03/16) Equilibrium net = ma; system: diving board ext 1 + 2 + mg = ma 1 + 2 mg = 0 0, equilibrium 2 =? +y + Bolt 1 =? 3.9 m τ net = 0 ext τ 1 + τ 2 + τ = 0 g τ 1 + 0 τ = 0 g τ 1 = τ g l 1 1 = l w mg 1 = l w mg l 1 1.4 m ulcrum 2 = 1 + mg 2 = 946 N + 530N 2 = 1,476 N = 2.5m (530N) 1.4 m = 946 N g = mg =
Warm Up (02/04/16) Calculating Torque ou are installing a new spark plug in your car, and the manual specifies that it be tightened to a torque that has a magnitude of 45 N m. Using the data in the drawing, determine the magnitude of the force that you must exert on the wrench Torque, τ = (Magnitude of the force) (Lever rm) +y τ = l + 45 N m = (0.28m in50 ) 0.28 m = 45 N m = 210 N (0.28m in50 ) l 50.0 B