2/26/2014 Chapter 8: Torque and Angular Momentum Definition: rigid body rigid body, stationary axis of rotation rotational variables (and their similarity to translations): • angle: θ • angular velocity: ω = ∆θ/∆t • angular acceleration: α = ∆ω/∆t A B B A point at distance r from axis: relationship between v and ω rotations and kinetic energy Distance between any two points does not change rotational inertia Fixed axis of rotation Translation and Rotation For now, we consider only the case when the axis of rotation, wherever it is, is stationary Rotation: Translation: all points rotate around all points move some axis by exactly and have exactly the the same angle same displacements axis of rotation x1 θ2 θ1 x2 x-axis Translation and Rotation Translation coordinate: x (meters) Radians Rotation angle: θ (radians) DO NOT USE DEGREES (many formulas will work only in radians) θ= axis of rotation r x1 θ2 180º = π rad s s r θ θ1 x2 s = arc length x-axis 1 2/26/2014 Kinematics Translation position: x velocity: v = ∆x/∆ ∆t acceleration: a = ∆v/∆ ∆t More equations of motion Rotation (Ch. 5) angle: θ angular velocity: ω = ∆θ θ/ ∆t angular acc.: α = ∆ω /∆t axis of rotation x1 θ2 Translation Rotation a = const v = vi + at α = const ω = ωi + αt 1 x = xi + vit + at 2 2 θ = θi + ωi t + αt 2 1 2 θ1 x2 x-axis Acceleration and Angular Acceleration Kinetic Energy and Moment of Inertia Ki = we already knew that! v2 ar = = rω 2 r v r θ s at 1 1 1 2 mi vi2 = mi ( ω ri ) = ( mi ri 2 ) ω 2 2 2 2 Rotational Inertia I - characterizes the rotational inertia of the object - depends on the axis ar v r I { at = r α K =∑ and this ! i Kinetic Energy and Rotational Inertia I = ∑ mi ri 2 Compare to linear motion: M = ∑ mi i 1 K = Iω 2 2 (pure rotation) K= 1 Mv 2 2 1 ( mi ri2 ) ω 2 = 12 ∑ mi ri2 ω 2 = 12 Iω 2 2 i piece i of rigid body, mass mi Example 1 In a flywheel in the form of a disc, the rotational KE is 2x108 J at peak speed of ω=460 rad/s. 1. How much KE does the wheel have at its minimum speed of 300 rad/s? 2. The flywheel has a mass of 20,000 kg. What is its radius? 3. Suppose the flywheel is slowed down from max to minimum speed in 0.2 s. How many kW of avg power would it produce during this interval? 2 2/26/2014 Example 1 In a flywheel in the form of a disc, the rotational KE is 2x108 J at peak speed of ω=460 rad/s. 1. How much KE does the wheel have at its minimum speed of 300 rad/s? 1 2 I ωmax 2 1 2 = I ωmin 2 Example 1 In a flywheel in the form of a disc, the rotational KE is 2x108 J at peak speed of ω=460 rad/s. 2. The flywheel has a mass of 20,000 kg. What is its radius? K max = K min K max ωmax = K min ωmin 2 K max = 1 1 1 2 2 I ωmax = ( MR 2 )ωmax 2 2 2 ⇒ R= 2 ⇒ K min = K max ωmin 2 3002 = (2 × 108 J) = 8.5 ×107 J 460 2 ωmax 2 Example 1 2 ω K max 2 2 ×108 J = = 0.435m M 460 rad/s 2 ×104 kg Torque (Latin for “twist”) In a flywheel in the form of a disc, the rotational KE is 2x108 J at peak speed of ω=460 rad/s. 3. Suppose the flywheel is slowed down from max to minimum speed in 0.2 s. How many kW of avg. power would it produce during this interval? To open door, one needs to apply force The ease of opening depends on • distance between the force and the hinge • the direction of the force F ∆E 2 ×108 J-8.5 ×107 J P= = = 5.75 ×105 kW ∆t 0.2s Torque (Latin for “twist”): τ = Fr sin θ axis r F θ To twist an object, one needs to apply force The ease of twisting will depend on • force • distance between the force and the axis • direction of the force Torque: two perspectives τ = Fr sin θ τ = ( F sin θ )r = F⊥ r τ = F (r sin θ ) = Fr⊥ line of action r⊥ =r sinθ θ axis axis lever arm r r F⊥ =F sinθ θ θ F θ F 3 2/26/2014 Torque: Torque and work θ2 θ axis s r s W = ( F sin φr ) r W = τθ W = ( F sin φ )r r r1 s Units: • N-m φ F sinφ φ θ1 F F1 P= torque negative Example Lever arm for F2 F2=30 N ∆t = τω 30° F3=20 N 30° 10° X 45° F3=20 N X τ∆θ Example continued: Example: Calculate the torque due to the three forces shown about the left end of the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the bar. F2=30 N Note analogy with W=Fs Proof for simple case of one particle W = ( F sin φ ) s axis r2 Multiple forces: • τnet = τ1+ τ2 + … W = τθ torque positive F2 Sign convention: • CW (clockwise): negative • CCW(counter-clockwise): positive 10° 45° F1=25 N F1=25 N Lever arm for F3 r1 = 0 The lever arms are: 22 Mechanical Energy of Rigid Body Example continued: The torques are: r2 = (2m )sin 60° = 1.73 m r3 = (4m ) sin 10° = 0.695 m 21 τ1 = 0 τ 2 = +(1.73 m )(30 N ) = +51.9 Nm τ 3 = +(0.695 m )(20 N ) = +13.9 Nm The mechanical energy of a rigid body with fixed axis: Emec = U + K rot Gravitational Potential Energy U = ∑ mi gyi = g ∑ mi yi = gyCM M The net torque is +65.8 Nm and is the sum of the above results. Conservation of Energy Work-Energy 23 4 2/26/2014 Exam 2 Fall 2011: Problem 13 • A thin stick with mass M, length L, and moment of inertia ML2/3 is hinged at its lower end and allowed to fall freely as shown in the figure. If its length L = 2 m and it starts from rest at an angle θ = 20o, what is the speed (in m/s) of the free end of the stick when it hits the table? Answer: 7.43 m/s % Right: 14% Ei = MgyCM = Mg vf = L cos θ 2 E f = 12 Iω 2f = 12 I v 2f L2 hinge θ h= L cos θ 2 Ei = E f mgL3 cos θ mgL3 cosθ = = 3 gL cos θ 2 1 I 3 mL = 3(9.8m / s 2 )(2m) cos(20o ) ≈ 7.43m / s University of Florida PHY 2053 Page 25 5