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Chapter 8 Rotational Dynamics continued 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) Uniform circular motion centripetal acceleration centripetal force α = a / r (rad./s 2 ) v2 aC = = ω 2 r r mv 2 FC = maC = r I Moment of Inertia torque (θ : between F & r) Newton's 2nd Law rot. kinetic energy angular momentum τ = rF sin θ K rot = Iω L = Iω Moments of Inertia of Rigid Objects Mass M Thin walled hollow cylinder τ Net = I α 1 2 From Ch.8.4 Rotational Kinetic Energy I = MR 2 Solid sphere through center I = 25 MR 2 Solid sphere through surface tangent I = 75 MR 2 Thin walled sphere through center Solid cylinder or disk 2 I = MR 1 2 2 Thin rod length ! through center Thin plate width !, through center I = 121 M !2 I = 121 M !2 Thin rod length ! through end I = 13 M !2 I = 23 MR 2 Thin plate width ! through edge I = 13 M !2 Clicker Question 8.4 A weed wacker is a length of nylon string that rotates rapidly around one end. The rotation angular speed is 47.0 rev/s, and the tip has tangential speed of 54.0 m/s. What is length of the nylon string? ω a) 0.030 m b) 0.120 m c) 0.180 m d) 0.250 m e) 0.350 m L vT Clicker Question 8.5 The sun moves in circular orbit with a radius of 2.30x104 light-yrs, (1 light-yr = 9.50x1015 m) around the center of the galaxy at an angular speed of 1.10x10–15 rad/s. galaxy Find the tangential speed of sun. r a) 2.40 × 105 m/s b) 3.40 × 105 m/s c) 4.40 × 105 m/s d) 5.40 × 105 m/s e) 6.40 × 105 m/s ω sun Clicker Question 8.5 The sun moves in circular orbit with a radius of 2.30x104 light-yrs, (1 light-yr = 9.50x1015 m) around the center of the galaxy at an angular speed of 1.10x10–15 rad/s. galaxy Find the tangential speed of sun. r a) 2.40 × 105 m/s b) 3.40 × 105 m/s c) 4.40 × 105 m/s d) 5.40 × 105 m/s e) 6.40 × 105 m/s Clicker Question 8.6 Find the centripetal force on the sun. 30 20 m = 1.99 × 10 kg a) 2.27 × 10 N Sun ( b) 3.27 × 1020 N c) 4.27 × 1020 N d) 5.27 × 1020 N e) 6.27 × 1020 N ) ω sun 8.4 Rotational Work and Energy Total Energy = (Translational Kinetic + Rotational Kinetic + Potential) Energy E = 12 mv 2 + 12 Iω 2 + mgy ENERGY CONSERVATION 1 2 E f = E0 v0 = ω 0 = 0 y0 = h0 mv 2f + 12 Iω 2f + mgy f = 12 mv02 + 12 Iω 02 + mgy0 ω =v R 1 2 mv 2f + 12 I v 2f R 2 = mgh0 v 2f (m + I R 2 ) = 2mgh0 vf = 2mgh0 = 2 m+ I R yf = 0 2gh0 1+ I mR 2 The cylinder with the smaller moment of inertia will have a greater final translational speed. I = mR 2 I = 12 mR 2 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 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 Rotational/Linear Dynamics Summary linear rotational x displacement θ velocity v ω a acceleration α m F inertia force/torque I = mr 2 τ = Fr sin θ Potential Energies U G = mgy or U G = −GM E m/RE 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 = τθ K rot = 12 Iω 2 Wrot ⇒ ΔK rot L = Iω τΔt = ΔL Conservation laws Conserved If WNC = 0, If Fext = 0, If τ = 0, ext quantity: E = K + U Psystem = ∑ p L = Iω 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. 8.7 Rigid Objects in Equilibrium Reasoning Strategy 1. Select the object to which the equations for equilibrium are to be applied. 2. Draw a free-body diagram that shows all of the external forces acting on the object. 3. Choose a convenient set of x, y axes and resolve all forces into components that lie along these axes. 4. Apply the equations that specify the balance of forces at equilibrium. (Set the net force in the x and y directions equal to zero.) 5. Select a convenient axis of rotation. Set the sum of the torques about this axis equal to zero. 6. Solve the equations for the desired unknown quantities. Clicker Question 8.7 A 5-kg ball and a 1-kg ball are positioned a distance L apart on a bar of negligible mass. How far from the 5-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 = 5kg 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 8.9 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° θ b) 28.7° c) 33.7° d) 38.7° e) 43.7° F1 F2 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? 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 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 friction with ground 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⊥ Example: A flywheel has a mass of 13.0 kg and a radius of 0.300m. What angular velocity gives it an energy of 1.20 x 109 J ? K = 12 Iω 2 ; I disk = 12 MR 2 = 1 2 ( 1 2 ) MR 2 ω 2 4K 4.80 × 109 J ω = = 2 MR (13.0kg)(0.300m)2 2 ω = 4.10 × 109 = 6.40 × 104 rad/s = 6.40 × 104 rad/s ( rev/(2π )rad ) = 1.02 × 104 rev/s ( 60s/min ) = 6.12 × 105 rpm