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Chapter 8 Rotational Dynamics conclusion 8.7 Center of Gravity DEFINITION OF CENTER OF GRAVITY The center of gravity of a rigid body is the point at which its weight can be considered to act when the torque due to the weight is being calculated. 8.7 Center of Gravity When an object has a symmetrical shape and its weight is distributed uniformly, the center of gravity lies at its geometrical center. 8.7 Center of Gravity General Form of xcg Balance point is under xcg Center of Gravity, xcg , for 2 masses W1x1 + W2 x2 xcg = W1 + W2 8.7 Center of Gravity Example: The Center of Gravity of an Arm The horizontal arm is composed of three parts: the upper arm (17 N), the lower arm (11 N), and the hand (4.2 N). Find the center of gravity of the arm relative to the shoulder joint. W1x1 + W2 x2 + W3 x3 xcg = W1 + W2 + W3 ⎡⎣17 ( 0.13) + 11( 0.38) + 4.2 ( 0.61) ⎤⎦ N ⋅ m = = 0.28 m (17 + 11+ 4.2) N 8.7 Center of Gravity Finding the center of gravity of an irregular shape. Clicker Question 8.7 Torque and Equilibrium A 4-kg ball and a 1-kg ball are positioned a distance L apart on a bar of negligible mass. How far from the 4-kg mass should the fulcum be placed to balance the bar? L a) 1 2 L b) 1 3 L c) 1 4 L d) 1 5 L e) 1 6 L m1 = 1kg m2 = 4kg Clicker Question 8.7 Torque and Equilibrium A 4-kg ball and a 1-kg ball are positioned a distance L apart on a bar of negligible mass. How far from the 4-kg mass should the fulcum be placed to balance the bar? L a) 1 2 L m1 = 1kg b) 1 3 L positive torque c) 1 4 L d) 1 5 L e) 1 6 L 4 W1 1 m2 = 4kg negative torque W2 For equilibrium the sum of the torques must be zero Need to separate length into 4 parts on 1-kg mass side and 1 part on the 4-kg mass side. Total is 5 parts. Fulcum must be 1/5 of the total length from the 4-kg mass. Clicker Question 8.7 Torque and Equilibrium A 4-kg ball and a 1-kg ball are positioned a distance L apart on a bar of negligible mass. How far from the 4-kg mass should the fulcum be placed to balance the bar? L a) 1 2 L m1 = 1kg b) 1 3 L positive torque c) 1 4 L d) 1 5 L e) 1 6 L L− x x m2 = 4kg negative torque W1 W2 For equilibrium the sum of the torques must be zero Let x be the distance of fulcum from 4-kg mass. ∑τ = 0 = m g(L − x) + (−m gx) (m + m ) x = m L 1 2 1 2 1 x= m1 1 1 L= L= L ( m1 + m2 ) (1+ 4) 5 8.7 Rigid Objects in Equilibrium EQUILIBRIUM OF A RIGID BODY A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero. Note: constant linear speed or constant rotational speed are allowed for an object in equilibrium. Clicker Question 8.8 A 0.280 m long wrench at an angle of 50.0° to the floor is used to turn a nut. What horizontal force F produces a torque of 45.0 Nm on the nut? L = 0.280m a) 10 N b) 110 N c) 210 N d) 310 N e) 410 N θ = 50° F Clicker Question 8.8 A 0.280 m long wrench at an angle of 50.0° to the floor is used to turn a nut. What horizontal force F produces a torque of 45.0 Nm on the nut? L = 0.280m a) 10 N F⊥ θ = 50° b) 110 N θ c) 210 N F d) 310 N τ = F⊥ L = ( F sin θ ) L e) 410 N F = τ Lsin θ = 45Nm (0.28 m)sin50° = 210 N Clicker Question 8.9 A force F1 = 38.0 N acts downward on a horizontal arm. A second force, F2 = 55.0 N acts at the same point but upward at the angle theta. What is the angle of F2 for a net torque = zero? a) 23.7° b) 28.7° c) 33.7° d) 38.7° e) 43.7° θ F2 = 55.0N F1 = 38.0N Clicker Question 8.9 A force F1 = 38.0 N acts downward on a horizontal arm. A second force, F2 = 55.0 N acts at the same point but upward at the angle theta. What is the angle of F2 for a net torque = zero? a) 23.7° F2 sin θ b) 28.7° c) 33.7° d) 38.7° e) 43.7° θ F1 τ Net = τ 1 + τ 2 = 0 (τ 1 is clockwise ⇒ –) τ Net = − F1 L + ( F2 sin θ ) L = 0 sin θ = F1 F2 = 38.0 / 55.0 = 0.691 θ = 43.7° F2 8.7 Rigid Objects in Equilibrium Example: A board length L lies against a wall. The coefficient of friction with ground is 0.650. What is smallest angle the board can be placed without slipping? 1. Determine the forces acting on the board. θ 8.7 Rigid Objects in Equilibrium Example: A board length L lies against a wall. The coefficient of friction with ground is 0.650. What is smallest angle the board can be placed without slipping? θ 1. Determine the forces acting on the board. Forces: Forces P G y ground normal force Gy = W Gx ground static frictional force P wall normal force W gravitational force L/2 Gy θ Gx 2. Choose pivot point at ground. W P = Gx = µG y = µW 8.7 Rigid Objects in Equilibrium Example: A board length L lies against a wall. The coefficient of static friction with ground is 0.650. What is smallest angle the board can be placed without slipping? θ 1. Determine the forces acting on the board. Forces: Forces P G y ground normal force Gy = W Gx ground static frictional force P wall normal force W gravitational force L/2 Gy W P = Gx = µG y = µW θ P⊥ P Gx 2. Choose pivot point at ground. 3. Find components normal to board P⊥ and W⊥ are forces producing torque L/2 Gy θ Gx W W⊥ 8.7 Rigid Objects in Equilibrium Example: A board length L lies against a wall. The coefficient of friction with ground 0.650. What is smallest angle the board can be placed without slipping? 4. Net torque must be zero for equilibrium Torque: τ P = + P⊥ L = ( Psin θ ) L τ W = −W⊥ ( L / 2 ) = − (W cosθ ) ( L / 2 ) Forces: Gy = W P = Gx = µG y = µW τ W + τ P = 0 ⇒ ( Psin θ ) = (W cosθ ) / 2 W = 2Psin θ / cosθ θ P⊥ P L/2 θ W⊥ 8.7 Rigid Objects in Equilibrium Example: A board length L lies against a wall. The coefficient of friction with ground 0.650. What is smallest angle the board can be placed without slipping? 4. Net torque must be zero for equilibrium Torque: τ P = + P⊥ L = ( Psin θ ) L τ W = −W⊥ ( L / 2 ) = − (W cosθ ) ( L / 2 ) Forces: Gy = W P = Gx = µG y = µW τ W + τ P = 0 ⇒ ( Psin θ ) = (W cosθ ) / 2 W = 2Psin θ / cosθ 5. Combine torque and force equations W = 2Psin θ / cosθ = 2 µW tan θ tan θ = 1 (2 µ ) = 1 (1.3) = 0.77 θ = 37.6° θ P⊥ P L/2 θ W⊥ Chapter 8 Rotational Kinematics/Dynamics Angular motion variables (with the usual motion equations) displacement velocity acceleration θ = s / r (rad.) ω = v / r (rad./s) α = a / r (rad./s 2 ) rot. kinetic energy K rot = 12 Iω 2 torque τ = rF sin θ (θ : between F & r) work rotating by φ Wrot = τφ Uniform circular motion centripetal acceleration centripetal force v2 aC = = ω 2 r r mv 2 FC = maC = r I Moment of Inertia 8.8 Newton’s Second Law for Rotational Motion About a Fixed Axis τ = FT r = maT r = mα r 2 ( ) = mr α 2 = Iα I = mr 2 Moment of Inertia of the airplane Moment of Inertia, I =mr2, for a point-mass, m, at the end of a massless arm of length, r. τ = Iα Newton’s 2nd Law for rotations 8.8 Newton’s Second Law for Rotational Motion About a Fixed Axis Break object into N individual masses τ Net = ∑ (mr )i α 2 i Net external torque τ2 Moment of inertia ( ) = (m r )α τ 1 = m1r12 α 2 2 2 2 τ N = mN rN α ( ) 8.8 Newton’s Second Law for Rotational Motion About a Fixed Axis ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS τ Net = I α Requirement: Angular acceleration must be expressed in radians/s2. I = ∑ (mr 2 ) i i 8.8 Newton’s Second Law for Rotational Motion About a Fixed Axis dual pulley Example: Hoisting a Crate The combined moment of inertia of the dual pulley is 50.0 kg·m2. The crate weighs 4420 N. A tension of 2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual Pulley (radius-1 = 0.600m, radius-2 = 0.200 m). Tension 1 Dual Pulley Motor Tension 2 up is positive Forces on Pulley P Post 1 = 0.600m T1 Axis WP 2 = 0.200m Forces on Crate T2′ ay T2 Crate W magnitude mg 8.8 Newton’s Second Law for Rotational Motion About a Fixed Axis 2nd law for linear motion of crate ∑F y = T2 − mg = ma y a y = 2α 2nd law for rotation of the pulley I = 46kg ⋅ m 2 mg = 4420 N T1 = 2150 N P 1 = 0.600m Dual Pulley T1 2 = 0.200m Motor Axis T2 − mg = m 2α T11 − T2 2 = Iα T2 = m 2α + mg & T2 = T11 − Iα 2 m22α + Iα = T11 − mg 2 α= WP T2 = T2 ,down T2′ = T2 ,up (+) Crate T11 − mg 2 2 = 6.3 rad/s m22 + I ay T2 = m 2α + mg = ( 451(.6)(6.3) + 4420 ) N = 6125 N mg 8.9 Angular Momentum DEFINITION OF ANGULAR MOMENTUM The angular momentum L of a body rotating about a fixed axis is the product of the body’s moment of inertia and its angular velocity with respect to that axis: Requirement: The angular speed must be expressed in rad/s. SI Unit of Angular Momentum: kg·m2/s 8.9 Angular Momentum PRINCIPLE OF CONSERVATION OF ANGULAR MOMENTUM The angular momentum of a system remains constant (is conserved) if the net external torque acting on the system is zero. Angular momentum, L Li = I iω i ; L f = I f ω f ⇒ No external torque ⇒ Angular momentum conserved L f = Li I f ω f = I iω i I = ∑ mr 2 , r f < ri Moment of Inertia decreases I f < Ii Ii >1 If Ii ω f = ωi; If Ii >1 If ω f > ω i (angular speed increases) 8.9 Angular Momentum From Angular Momentum Conservation ( ) ω f = Ii I f ω i ⇒ because I i I f > 1 Angular velocity increases Is Energy conserved? K f = 12 I f ω 2f = I f Ii I f ( ) = Ii I f I iω 2i 1 2 2 ω 2i ( )( ) = (I I )K ⇒ i f 1 2 i K i = 12 I iω 2i ; Kinetic Energy increases Energy is NOT conserved because pulling in the arms does (NC) work on the mass of each arm and increases the kinetic energy of rotation. 8.9 Angular Momentum Example: A Satellite in an Elliptical Orbit An artificial satellite is placed in an elliptical orbit about the earth. Its point of closest approach is 8.37x106 m from the center of the earth, and its point of greatest distance is 25.1x106 m from the center of the earth.The speed of the satellite at the perigee is 8450 m/s. Find the speed at the apogee. I A = mrA2 ; I P = mrP2 ω A = vA rA ; ω P = vP rP Gravitational force is along r (no torque) ⇒ Angular momentum conserved I Aω A = I Pω P mrA2 ( vA rA ) = mrP2 ( vP rP ) ⇒ rA vA = rP vP vA = ( rP rA ) vP = ⎡⎣(8.37 × 106 ) / (25.1× 106 ) ⎤⎦ (8450 m/s ) = 2820 m/s 8.9 The Vector Nature of Angular Variables Right-Hand Rule: Grasp the axis of rotation with your right hand, so that your fingers circle the axis in the same sense as the rotation. ω Your extended thumb points along the axis in the direction of the angular velocity. Vector Quantities in Rotational Motion Angular Acceleration Angular Momentum Torque Δω α= Δt L = Iω ΔL τ=Iα= Δt ω Changes Angular Momentum If torque is perpendicular to the angular momentum, only the direction of the angular momentum changes (precession) – no changes to the magnitude. Rotational/Linear Dynamics Summary linear rotational x displacement θ velocity v ω a acceleration α m F point m inertia force/torque I = mr 2 τ = Fr sin θ Uniform circular motion aC = vT2 r FC = mvT2 r Potential Energies Gravitation GmM U G = mgy (on earth) F= 2 r or U G = −GmM E /r U S = 12 kx 2 linear F = ma W = F(cosθ )x K = 12 mv 2 W ⇒ ΔK p = mv FΔt = Δp rotational τ = Iα Wrot = τφ K rot = 12 Iω 2 Wrot ⇒ ΔK rot L = Iω τΔt = ΔL Conservation laws If WNC = 0, If Fext = 0, Conserved ⇒ E = K + U Psystem If τ ext = 0, = ∑ p L = Iω