Ch 02 modeling of rigid body mechanical systems

8/25/2013 System Dynamics 2.01 Modeling of Rigid Body Mechanical Systems Modeling of Rigid Body Mechanical Systems HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.03 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §1.Translational Motion 1.Mechanical Energy - Kinetic Energy: the work needed to accelerate a body of a given mass from rest to its stated velocity 𝑑𝑣 𝑚𝑣 = 𝑚 = 𝑓(𝑥) 𝑑𝑡 𝑑𝑣 ⟹ 𝑣𝑚 = 𝑣𝑓(𝑥) 𝑑𝑡 ⟹ 𝑚𝑣𝑑𝑣 = 𝑣𝑓 𝑥 𝑑𝑡 = 𝑓 𝑥 𝑑𝑥 ⟹ 𝑚𝑣𝑑𝑣 = 𝑚𝑣 = 𝑓 𝑥 𝑑𝑥 + 𝐶 Work is force times displacement, so the integral on the right represents the total work done on the mass by the force 𝑓(𝑥) 𝐾𝐸 = 𝑚𝑣 2 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.05 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems System Dynamics 2.02 Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Particle: a mass of negligible dimensions A body is to be a particle if its dimensions are irrelevant for specifying its position and the forces acting on it - Newton’s Second Law 𝑑𝒗 𝑚𝒂 = 𝑚 =𝑭 𝑑𝑡 𝑚: mass of particle 𝒂: acceleration of particle 𝒗: velocity of particle 𝑭: force acting on particle - If the mass is constrained to move in only 𝑥 direction, then the equation of motion is the scalar equation 𝑑𝑥 𝑚𝑥 = 𝑚 =𝑓 𝑑𝑡 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.04 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Potential Energy: the energy of an object or a system due to the position of the body or the arrangement of the particles of the system - Potential Energy Function 𝑉 𝑥 If the work done by a conservatie force 𝑓(𝑥) (independent of the path taken, depends only on the end points), then the force 𝑓(𝑥) is derivable from a function 𝑉(𝑥) as 𝑑𝑉(𝑥) 𝑓 𝑥 =− 𝑑𝑥 ⟹𝑉 𝑥 = 𝑑𝑉 𝑥 = − 𝑓 𝑥 𝑑𝑥 ⟹ 𝑚𝑣 + 𝑉 𝑥 = 𝐶 Sum of the kinetic and potential energies must be constant HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.06 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Gravity force: the gravity force is conservative because the work done lifting an object depends only on the change in height and not on the path taken 𝑓 𝑥 = −𝑚𝑔 𝑥 𝑥: vertical displacement Potential energy function 𝑉 𝑥 = 𝑚𝑔𝑥 The sum of energy 𝑚𝑣 + 𝑚𝑔𝑥 = 𝐶 ⟹ 𝑚(𝑣 − 𝑣02 ) + 𝑚𝑔(𝑥 − 𝑥0 ) = §1.Translational Motion - Example 2.1.1 Speed of a Falling Object An object with a mass of 𝑚 = 2𝑠𝑙𝑢𝑔𝑠 drops from a height of 30𝑓𝑡 above the ground Determine its speed after it drops to 10𝑓𝑡 above the ground 𝑔 = 32.2𝑓𝑡/𝑠𝑒𝑐 Solution From the energy conservation law 𝑚 𝑣 − 𝑣02 + 𝑚𝑔 𝑥 − 𝑥0 = HCM City Univ of Technology, Mechanical Engineering Department HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien ⟹𝑣= 2𝑔(𝑥0 − 𝑥) + 𝑣02 = × 32.2 30 − 10 + 02 𝑣 = 35.89𝑓𝑡/𝑠𝑒𝑐 Nguyen Tan Tien 8/25/2013 System Dynamics 2.07 Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Constant Force Case Energy conservation law 1 𝑚𝑣 + 𝑉 𝑥 = 𝑚𝑣02 + 𝑉 𝑥0 = 𝐶 2 For the point mass model, 𝑓 = 𝑚𝑎 If 𝑓 is a constant 1 𝑚𝑣 = 𝑓 𝑥 − 𝑥0 + 𝑚𝑣02 2 𝑓 𝑥 − 𝑥0 : the work done by the force 𝑓 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.09 System Dynamics 2.08 Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Dry Friction Force Dry friction is non-conservative force 𝐹 = 𝜇𝑁 Dry friction cannot be derived from a potential energy function ⟹ the conservation of mechanical energy principle does not apply Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Example 2.1.2 Equation of Motion with Friction Apply a force 𝑓1 to the mass 𝑚, derive the equation of motion (a) for the block of mass 𝑚 on an horizontal plane HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.10 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §1.Translational Motion Solution (a) The block of mass 𝑚 on an horizontal plane Dry friction force 𝐹 = 𝜇𝑚𝑔 𝑣>0 𝑚𝑣 = 𝑓1 − 𝜇𝑚𝑔 (b) for the mass 𝑚 on an incline plane 𝑣0 𝑚𝑣 = 𝑓1 − 𝑚𝑔𝑠𝑖𝑛𝜙 − 𝜇𝑚𝑔𝑐𝑜𝑠𝜙 𝑣 0: 𝑚𝑣 = 𝑓1 − 𝑚𝑔𝑠𝑖𝑛𝜙 − 𝜇𝑚𝑔𝑐𝑜𝑠𝜙 ⟹ 2𝑣 = 𝑓1 − (𝑠𝑖𝑛300 + 0.5𝑐𝑜𝑠300)(2 × 9.81) = 𝑓1 − 18.3 b 𝑣 = Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §1.Translational Motion - Example 2.1.3 Motion with Friction on an Inclined Plane For the mass 𝑚 = 2𝑘𝑔, 𝜙 = 300 , 𝑣(0) = 3𝑚/𝑠, and 𝜇 = 0.5 Determine whether the mass comes to rest if (a) 𝑓1 = 50𝑁 and (b) 𝑓1 = 5𝑁 a 𝑣 = HCM City Univ of Technology, Mechanical Engineering Department 2.12 Nguyen Tan Tien 50−18.3 5−18.3 = 15.85 > ⟹ the mass never comes to rest = −6.65 < ⟹ the mass will comes to rest HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.13 Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis - Newton’s Second Law 𝐼𝜔 = 𝑀 𝜔: the angular velocity of the mass 𝐼: the mass moment of inertia of the body 𝑀: the sum of the moments applied to the body System Dynamics 𝐼= System Dynamics 2.15 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis - Mass moments of inertia of common elements Sphere Hollow cylinder 𝐼𝐺 = 𝑚𝑅 𝐼𝑥 = 𝑚(𝑅 + 𝑟 ) 𝐼𝑦 = 𝐼𝑧 = 𝑚(3𝑅 + 3𝑟 + 𝐿2 ) 12 Mass rotating about 𝑂 Modeling of Rigid Body Mechanical Systems 𝑟 𝑑𝑚 𝐼: mass moment of inertia about a specified reference axis 𝑟: distance from the reference axis 𝑑𝑚: the mass element Parallel-axis theorem 𝐼 = 𝐼𝑠 + 𝑚𝑑2 𝐼𝑠 : HCM City Univ of Technology, Mechanical Engineering Department 2.14 §2.Rotation about Fixed Axis 1.Calculating Inertia Rectangular prism the inertia about the symmetry axis HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.16 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis - Example 2.2.1 A Rod-and-Bob Pendulum The pendulum consists of a concentrated mass 𝑚𝐶 a distance 𝐿𝐶 from point 𝑂 , attached to a rod of length 𝐿𝑅 and inertia 𝐼𝑅𝐺 about its mass center (a) Obtain its equation of motion (b) Discuss the case where the rod’s mass 𝑚𝐺 is small compared to the concentrated mass (c) Determine the equation of motion for small angles 𝜃 𝐼𝑂 = 𝑚𝑅 𝐼𝑥 = 𝑚(𝑏 + 𝑐 ) 12 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.17 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis Solution (a) Obtain its equation of motion The equation of motion 𝐼𝑂 𝜃 = −𝑚𝑔𝐿𝑠𝑖𝑛𝜃 Pendulum’s inertia about point 𝑂 𝐼𝑂 = 𝐼𝑅𝑂 + 𝑚𝐶 𝐿2𝐶 The rod’s inertia about point 𝑂 𝐿𝑅 𝐿𝑅 𝐼𝑅𝑂 = 𝐼𝑅𝐺 + 𝑚𝑅 ⟹ 𝐼𝑂 = 𝐼𝑅𝐺 + 𝑚𝐺 2 Taking moments about point 𝑂 to get the length 𝐿 𝑚𝑔𝐿 = 𝑚𝐶 𝑔𝐿𝐶 + 𝑚𝑅 𝑔 + 𝑚𝐶 𝐿2𝐶 𝐿 𝑚𝐶 𝐿𝐶 + 𝑚𝑅 𝑅 𝐿𝑅 ⟹𝐿= 𝑚𝐶 + 𝑚𝑅 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.18 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis (b) Discuss the case where the rod’s mass 𝑚𝐺 is small compared to the concentrated mass If neglect the rod’s mass 𝑚𝑅 compared to the concentrated mass 𝑚𝐶 𝑚𝑅 = 0; 𝐼𝑅𝐺 = 0; 𝑚 = 𝑚𝐶 ; 𝐿 = 𝐿𝐶 ; 𝐼𝑂 = 𝑚𝐿2 The equation of motion 𝐿𝜃 + 𝑔𝑠𝑖𝑛𝜃 = This is a model for a pendulum whose mass is concentrated at a distance 𝐿 form the pivot point This equation of motion is independent of the value of 𝑚 (c) For small angles, sin𝜃 ≈ 𝜃 𝐿𝜃 + 𝑔𝜃 = HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.19 Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis 2.Energy and Rotational Motion Work done by a moment 𝑀 causing a rotation through an angle 𝜃 𝜃 𝑊= 𝑀𝑑𝜃 System Dynamics 2.20 Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis 3.Pulley Dynamics - Pulleys can be used to change the direction of an applied force or to amplify forces 𝐼𝜃 = 𝑅𝑇1 − 𝑅𝑇2 = 𝑅(𝑇1 − 𝑇2 ) Multiply both sides of 𝐼𝜔 = 𝑀 ⟹ 𝐼𝑑𝜔 = 𝑀𝑑𝑡 ⟹ 𝐼𝜔𝑑𝜔 = 𝑀𝜔𝑑𝑡 = 𝑀𝑑𝜃 Integrating both sides 𝜔 𝜃 𝐼𝜔𝑑𝜔 = 𝐼𝜔2 = 𝑀𝑑𝜃 0 𝐼: moment of inertia of the pulley 𝑅: pulley radius 𝑇𝑖 : the tension forces acting on belt 𝑀 produces the kinetic energy of rotation 𝐾𝐸 = 𝐼𝜔2 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.21 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis - Example 2.2.2 Energy Analysis of a Pulley System Consider the pulley system used to raise the mass 𝑚2 by hanging a mass 𝑚1 on the other side of the pulley If pulley inertia is negligible then it is obvious that 𝑚1 will lift 𝑚2 if 𝑚1 > 𝑚2 - How does a non-negligible pulley inertia 𝐼 change this result? - Investigate the effect of the pulley inertia on the speed of the masses HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.22 System Dynamics 2.23 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis - Example 2.2.3 Equation of Motion of a Pulley System Consider the pulley system Obtain the equation of motion in terms of 𝑥 and obtain an expression for the tension forces in the cable Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis Solution Define the coordinates 𝑥 and 𝑦 such that 𝑥 = 𝑦 = at the start of the motion Pulley cable is inextensible 𝑥 = 𝑦, 𝑥 = 𝑦 Cable does not slip 𝜃 = 𝑥/𝑅 Kinetic energy 1 2 𝐾𝐸 = 𝑚1 𝑥 + 𝑚2 𝑦 + 𝐼 𝜃 Potential energy 𝑃𝐸 = 𝑚2 𝑔𝑦 − 𝑚1 𝑔𝑥 Energy conservation law 𝐾𝐸 + 𝑃𝐸 = 𝐼 ⟹ 𝑚1 + 𝑚2 + 𝑥 + 𝑚2 − 𝑚1 𝑔𝑥 = 𝑅 ⟹𝑥= HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 𝑚2 − 𝑚1 𝑚1 + 𝑚2 + 𝐼/𝑅2 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.24 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis Solution Newton’s law for masses 𝑚1 and 𝑚2 gives 𝑚1 𝑥 = 𝑚1 𝑔 − 𝑇1 𝑇 = 𝑚1 𝑔 − 𝑚1 𝑥 ⟹ 𝑚2 𝑦 = 𝑇2 − 𝑚2 𝑔 𝑇2 = 𝑚2 𝑦 + 𝑚1 𝑔 Newton’s law for the pulley 𝐼𝜃 = 𝑅𝑇1 − 𝑅𝑇2 Because 𝑥 = 𝑅𝜃, 𝑥 = 𝑅𝜃 𝑥 𝐼 = 𝑚1 − 𝑚2 𝑔𝑅 − 𝑚1 + 𝑚2 𝑅𝑥 𝑅 𝑚1 − 𝑚2 𝑔𝑅 ⟹𝑥= 𝑚1 + 𝑚2 𝑅 + 𝐼 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.25 Modeling of Rigid Body Mechanical Systems System Dynamics 2.26 §2.Rotation about Fixed Axis 4.Pulley-Cable Kinematics - Determine the relation between the velocities of mass 𝑚𝐴 and mass 𝑚𝐵 • Define 𝑥 and 𝑦 as shown from a common reference line attached to a fixed part of the system • Noting that the cable lengths wrapped around the pulleys are constant 𝑥 + 3𝑦 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ⟹ 𝑥 + 3𝑦 = (the speed of point 𝐴 is three times the speed of point 𝐵, and in the opposite direction) §2.Rotation about Fixed Axis - Example 2.2.4 HCM City Univ of Technology, Mechanical Engineering Department HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.27 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis Solution Lifting a Mast A mast weighing 500𝑙𝑏 is hinged at its bottom to a fixed support at point 𝑂 The mast is 70𝑓𝑡 long and its center of mass is 35𝑓𝑡 from 𝑂 The winch applies a force 𝑓 = 380𝑙𝑏 to the cable The mast is supported initially at the 300 angle, and the cable at 𝐴 is initially horizontal Assume that the pulley inertias are negligible and that the pulley diameter 𝑑 is very small compared to the other dimensions Derive the equation of motion of the mast System Dynamics 𝑄= 𝑃2 + 𝐿2 − 2𝑃𝐿𝑐𝑜𝑠(𝜋 − 𝜇 − 𝜃) = 𝑃2 + 𝐿2 − 2𝑃𝐿𝑐𝑜𝑠(𝜇 + 𝜃) 𝐻 = 20𝑓𝑡 𝑊 = 5𝑓𝑡 geometry of the mast 𝜃 𝐻 + 𝑊 = 20.6𝑓𝑡 𝐻 10 𝜇 = 𝑡𝑎𝑛−1 = 𝑡𝑎𝑛−1 = 1.33 𝑊 𝑃= HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.28 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §2.Rotation about Fixed Axis The law of sines 𝑠𝑖𝑛𝜙 sin(𝜋 − 𝜇 − 𝜃) = 𝑃 𝑄 sin(𝜇 + 𝜃) = 𝑄 The law of cosines geometry of the mast 𝜃 Modeling of Rigid Body Mechanical Systems 2.29 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems The moment eq about 𝑂 𝑓𝑙𝑃 𝐼𝑂 𝜃 = −𝑚𝑔𝑅𝑐𝑜𝑠𝜃 + sin(𝜇 + 𝜃) 𝑄 The moment of inertia 1 500 𝐼𝑂 = 𝑚(70)2 = (70)2 3 32.2 = 25,400𝑠𝑙𝑢𝑔 − 𝑓𝑡 Force at 𝐴 𝑓 = × 380 = 760𝑙𝑏 The equation of motion 25,400𝜃 = −17,500𝑐𝑜𝑠𝜃 626,000 + sin(1.33 + 𝜃) 𝑄 𝑄 = 2020 + 1650cos(1.33 + 𝜃) HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.30 §3.Equivalent Mass and Inertia - Some systems composed of translating and rotating parts whose motions are directly coupled can be modeled as a purely translational system or as a purely rotational system, by using the concepts of equivalent mass and equivalent inertia These models can be derived using kinetic energy equivalence - Example 2.3.1 A Vehicle on an Incline: Energy Analysis A tractor pulls a cart up a slope, starting from rest and accelerating to 20𝑚/𝑠 The force in the cable is 𝑓, and the body of the cart has a mass 𝑚 The cart has two identical wheels, with radius 𝑅, mass 𝑚𝑤 , and inertia 𝐼𝑤 The two wheels are coupled with an axle whose mass is negligible Assume that the wheels not slip or bounce Derive an expression for the force 𝑓 using kinetic energy equivalence §3.Equivalent Mass and Inertia Solution Kinetic energy of the system HCM City Univ of Technology, Mechanical Engineering Department HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems 1 𝐾𝐸 = 𝑚𝑣 + 2𝑚𝑤 𝑣 + 2𝐼𝑤 𝜔2 2 𝐼𝑤 = 𝑚 + 2𝑚𝑤 + 2 𝑣 2 𝑅 Equivalent kinetic energy 𝐾𝐸 = 𝑚𝑒 𝑣 2 Equivalent mass 𝐼𝑤 𝑚𝑒 = 𝑚 + 2𝑚𝑤 + 2 𝑅 Nguyen Tan Tien 8/25/2013 System Dynamics 2.31 Modeling of Rigid Body Mechanical Systems System Dynamics 2.32 Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia Consider the free body diagram The potential energy of the system 𝑃𝐸 = 𝑚 + 2𝑚𝑤 𝑔𝑥𝑠𝑖𝑛𝜃 The motion equation 𝑚𝑒 𝑣 = 𝑓 − 𝑚 + 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 The acceleration 20 𝑎=𝑣= = 𝑚/𝑠 15 The force required to provide the specified acceleration 𝑓 = 𝑚𝑒 + 𝑚 + 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 §3.Equivalent Mass and Inertia 1.Mechanical Drivers Gears, belts, levers, and pulleys transform an input motion, force, or torque into another motion, force, or torque at the output - Example 2.3.2 Equivalent Inertia of Spur Gears Consider the spur gears Derive the expression for the equivalent inertia 𝐼𝑒 felt on the input shaft Solution 1 1 𝜔1 𝐾𝐸 = 𝐼1 𝜔12 + 𝐼2 𝜔22 = 𝐼1 𝜔12 + 𝐼2 2 2 𝑁 1 = 𝐼1 + 𝐼2 𝜔12 𝑁 HCM City Univ of Technology, Mechanical Engineering Department HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.33 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia - Example 2.3.3 Equivalent Inertia of a Rack-and-Pinion A rack-and-pinion is used to convert rotation into translation The input shaft rotates through the angle 𝜃 as a result of the torque 𝑇 produced by a motor The pinion rotates and causes the rack to translate Derive the expression for the equivalent inertia 𝐼𝑒 felt on the input shaft The mass of the rack is 𝑚, the inertia of the pinion is 𝐼, and its mean radius is 𝑅 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.35 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia - Example 2.3.4 Equivalent Inertia of a Belt Drive Belt drives and chain drives have similar characteristics and can be analyzed in a similar way The input shaft (shaft 1) is connected to a device that produces a torque 𝑇1 at a speed 𝜔1, and drives the output shaft (shaft 2) The mean sprocket radii are 𝑟1 and 𝑟2 , and their inertias are 𝐼1 and 𝐼2 The belt mass is 𝑚 Derive the expression for the equivalent inertia 𝐼𝑒 felt on the input shaft Equivalent inertia felt on the input shaft 𝐼𝑒 = 𝐼1 + 𝐼2 𝑁 System Dynamics 2.34 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia Solution Kinetic energy of the system 1 𝐾𝐸 = 𝑚𝑥 + 𝐼 𝜃 2 1 = 𝑚(𝑅𝜃)2 + 𝐼𝜃 2 = (𝑚𝑅 + 𝐼)𝜃 2 The equivalent inertia felt on the shaft 𝐼𝑒 = 𝑚𝑅 + 𝐼 The model of the system dynamics 𝐼𝑒 𝜃 = 𝑇 or can be expressed in term of 𝑥 𝐼𝑒 𝑥 = 𝑅𝑇 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.36 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia Solution Kinetic energy of the system 1 𝐾𝐸 = 𝐼1 𝜔12 + 𝐼2 𝜔22 + 𝑚𝑣 2 2 If the belt does not stretch, translational speed of the belt 𝑣 = 𝑟1𝜔1 = 𝑟2 𝜔2 Kinetic energy 1 2 𝐾𝐸 = 𝐼1 𝜔12 + 𝐼2 𝑟1 𝜔1 𝑟2 + 𝑚 𝑟1 𝜔1 2 = 𝐼1 + 𝐼2 𝑟1 𝑟2 the + 𝑚𝑟12 𝜔12 The equivalent inertia felt on the input shaft 𝐼𝑒 = 𝐼1 + 𝐼2 𝑟1 𝑟2 + 𝑚𝑟12 The dynamics of the system 𝐼𝑒 𝜔1 = 𝑇1 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.37 Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia - Example 2.3.4 A Robot-Arm Link A single link of a robot arm has mass 𝑚 and its center of mass is located a distance 𝐿 from the joint, which is driven by a motor torque 𝑇𝑚 through a pair of spur gears 𝑚 and 𝐿 depend on the payload being carried in the hand and thus can be different for each application The gear ratio is 𝑁 = The given values for the motor, shaft, and gear inertias are 𝐼𝑚 = 0.05𝑘𝑔𝑚2 , 𝐼𝐺1 = 0.025𝑘𝑔𝑚2 , 𝐼𝑆1 = 0.01𝑘𝑔𝑚2 , 𝐼𝐺2 = 0.1𝑘𝑔𝑚2 , 𝐼𝑆2 = 0.02𝑘𝑔𝑚2 Obtain this equation in terms of the angle 𝜃 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.39 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §3.Equivalent Mass and Inertia Equivalent inertia referenced to the motor shaft 𝐼𝑒 = 𝐼𝑚 + 𝐼𝑆1 + 𝐼𝐺1 + 𝐼𝑆2 + 𝐼𝐺2 + 𝑚𝐿2 = 0.115 + 0.25𝑚𝐿2 Equation of motion for this equivalent inertia 𝐼𝑒 𝜔1 = 𝑇𝑚 − 𝑚𝑔𝐿𝑠𝑖𝑛𝜃 𝑁 with 𝜔1 = 𝑁𝜔2 = 𝑁𝜃 𝐼𝑒 𝑁𝜃 = 𝑇𝑚 − 𝑚𝑔𝐿𝑠𝑖𝑛𝜃 𝑁 9.8 ⟹ 0.115 + 0.25𝑚𝐿2 𝜃 = 𝑇𝑚 − 𝑚𝐿𝑠𝑖𝑛𝜃 The equation of motion 0.23 + 0.5𝑚𝐿2 𝜃 = 𝑇𝑚 − 4.9𝑚𝐿𝑠𝑖𝑛𝜃 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.41 2.38 Modeling of Rigid Body Mechanical Systems Kinetic energy of the system 1 𝐾𝐸 = 𝐼𝑚 + 𝐼𝑆1 + 𝐼𝐺1 𝜔12 + 𝐼𝑆2 + 𝐼𝐺2 + 𝑚𝐿2 𝜔22 2 with 𝜔2 = 𝜔1 /𝑁 = 𝜔1/2 1 𝐾𝐸 = 𝐼𝑚 + 𝐼𝑆1 + 𝐼𝐺1 + 𝐼𝑆2 + 𝐼𝐺2 + 𝑚𝐿2 𝜔12 2 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.40 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion 1.Force Equations Two force equations translational motion 𝑓𝑥 = 𝑚𝑎𝐺𝑥 𝑓𝑦 = 𝑚𝑎𝐺𝑦 𝑓𝑥 , 𝑓𝑦 : describe the net forces acting on the mass 𝑚: mass 𝑎𝐺𝑥 , 𝑎𝐺𝑦 : accelerations of the mass center in the 𝑥, 𝑦 directions Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion 2.Moment Equations 𝐼𝑂 𝛼 = 𝑀𝑂 𝐼𝑂 : the mass moment of inertia of the body about the point 𝑂 𝛼: the angular acceleration of the mass 𝑀𝑂 : the sum of the moments applied to the body about the point 𝑀𝐺 = 𝐼𝐺 𝛼 𝑀𝐺 : net moment acting on the body about an axis that passes through the mass center 𝐺 and is perpendicular to the plane of the slab 𝐼𝐺 : the mass moment of inertia of the body about this axis 𝛼: the angular acceleration of the mass HCM City Univ of Technology, Mechanical Engineering Department System Dynamics §3.Equivalent Mass and Inertia Solution Nguyen Tan Tien HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.42 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion - Moment equation applies for an accelerating point 𝑃 𝑀𝑃 = 𝐼𝐺 𝛼 + 𝑚𝑎𝐺 𝑑 𝑀𝑃 : net moment acting on the body about an axis that passes through 𝑃 and is perpendicular to the plane of the slab 𝐼𝑃 : the mass moment of inertia of the body about this axis 𝛼: the angular acceleration of the mass 𝑚: mass 𝑎𝐺 : the magnitude of the acceleration vector 𝒂𝐺 𝑑: the distance between 𝒂𝐺 and a parallel line through point 𝑃 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.43 Modeling of Rigid Body Mechanical Systems §4.General Planar Motion - An alternative form 𝑀𝑃 = 𝐼𝑃 𝛼 + 𝑚𝑟𝑥 𝑎𝑃𝑦 − 𝑚𝑟𝑦 𝑎𝑃𝑥 𝑟𝑥 , 𝑟𝑦 : the 𝑥 , 𝑦 components of the location 𝐺 relative to 𝑃 𝑎𝑃𝑥 , 𝑎𝑃𝑦 : the 𝑥 , 𝑦 components of the acceleration of point 𝑃 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.45 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems System Dynamics 2.44 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.46 §4.General Planar Motion - Example 2.4.1 A Vehicle on an Incline: Force Analysis Consider a tractor pulls a cart up a slope The force in the cable is 𝑓, and the body of the cart has a mass 𝑚 The cart has two identical wheels, each with radius 𝑅, mass 𝑚𝑤 , inertia 𝐼𝑤 about the wheel center Assumed that the wheels not slip Develop a model that can be used to examine this assumption and also to compute the forces on the axle Apply Newton’s laws to develop a model of the system (a) Assume no slip and no bounce, show that the model gives the same result as of Example 2.3.1 (b) Use the model to discuss the no-slip assumption §4.General Planar Motion Solution HCM City Univ of Technology, Mechanical Engineering Department HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.47 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems a) Apply the Newton’s second law to the cart 𝑚𝑥 = 𝑅𝑥 − 𝑚𝑔𝑠𝑖𝑛𝜃 + 𝑓 𝑚𝑦 = 𝑅𝑦 − 𝑚𝑔𝑐𝑜𝑠𝜃 Apply the Newton’s second law to the wheel-axle subsystem 2𝑚𝑤 𝑥 = −𝑅𝑥 − 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 − 𝑓𝑡 2𝑚𝑤 𝑦 = 𝑁 − 𝑅𝑦 − 2𝑚𝑤 𝑔𝑐𝑜𝑠𝜃 Moment equation of the wheel-axle subsystem 2𝐼𝑤 2𝐼𝑤 𝑣 2𝐼𝑤 2𝐼𝑤 𝜔 = 𝑅𝑓𝑡 ⟹ 𝑓𝑡 = 𝜔= = 2𝑥 𝑅 𝑅 𝑅 𝑅 System Dynamics 2.48 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion Solve for 𝑅𝑥 𝑅𝑥 = − 2𝑚𝑤 + 2𝐼𝑤 𝑥 − 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 𝑅2 and 𝑚 + 2𝑚𝑤 + Modeling of Rigid Body Mechanical Systems §4.General Planar Motion 3.Sliding versus Rolling Motion - Wheels are common examples of systems undergoing general plane motion with both translation and rotation - Three possible motion types Pure rolling motion: no slipping between the wheel and the surface 𝑣 = 𝑅𝜔 Pure sliding motion: the wheel is prevented from rotating 𝜔 = 0, 𝑣 ≠ 𝑅𝜔 Sliding and rolling motion: slipping occurs in this case 𝜔 ≠ 0, 𝑣 ≠ 𝑅𝜔 2𝐼𝑤 𝑥 = 𝑓 − 𝑚 + 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 𝑅2 which the same as equation of Example 2.3.1 since 𝑥 = 𝑣 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien b) The assumption of no bounce ⟹ 𝑦 = 𝑁 = 𝑚 + 2𝑚𝑤 𝑔𝑐𝑜𝑠𝜃 No slip if 𝑓𝑡 ≤ 𝜇𝑠 𝑁 (the maximum static friction force) 2𝐼𝑤 2𝐼𝑤 𝑓 − 𝑚 + 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 𝑓𝑡 = 𝑥 = 2𝐼 𝑅 𝑅 𝑚 + 2𝑚𝑤 + 𝑤 𝑅2 ≤ 𝜇𝑠 𝑚 + 2𝑚𝑤 𝑔𝑐𝑜𝑠𝜃 𝑓 − 𝑚 + 2𝑚𝑤 𝑔𝑠𝑖𝑛𝜃 ⟹ 𝜇𝑠 𝑚 + 2𝑚𝑤 𝑔𝑐𝑜𝑠𝜃 ≥ 2𝐼𝑤 (𝑚 + 2𝑚𝑤 )𝑅 + 2𝐼𝑤 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.49 Modeling of Rigid Body Mechanical Systems §4.General Planar Motion - Example 2.4.2 A Rolling Cylinder A solid cylinder of mass 𝑚 and radius 𝑟 starts from rest and rolls down the incline at an angle 𝜃 The static friction coefficient is 𝜇𝑠 Determine the acceleration of the center of mass 𝑎𝐺𝑥 and the angular acceleration 𝛼 Assume that the cylinder rolls without bouncing, so that 𝑎𝐺𝑦 = Assume also that the cylinder rolls without slipping Use two approaches to solve the problem (a) Use the moment equation about 𝐺, and (b) Use the moment equation about 𝑃 (c) Obtain the frictional condition required for the cylinder to roll without slipping System Dynamics §4.General Planar Motion Solution 2.50 Modeling of Rigid Body Mechanical Systems Force equations 𝑚𝑎𝐺𝑥 = 𝑓𝑥 = 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝐹 𝑚𝑎𝐺𝑦 = 𝑓𝑦 = 𝑁 − 𝑚𝑔𝑐𝑜𝑠𝜃 Cylinder does not bounce 𝑎𝐺𝑦 = ⟹ 𝑁 = 𝑚𝑔𝑐𝑜𝑠𝜃 a Moment equation about the center of mass 𝐼𝐺 𝛼 = 𝑀𝐺 = 𝐹𝑟 𝐼𝐺 𝛼 𝑚𝑎𝐺𝑥 = 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝑟 The cylinder does not slip 𝑎𝐺𝑥 = 𝑟𝛼 𝐼𝐺 𝑎𝐺 𝑚𝑔𝑟 𝑠𝑖𝑛𝜃 𝑚𝑎𝐺𝑥 = 𝑚𝑔𝑠𝑖𝑛𝜃 − 𝑥 ⟹ 𝑎𝐺𝑥 = 𝑟 𝑚𝑟 + 𝐼𝐺 For a solid cylinder, 𝐼𝐺 = 𝑚𝑟 /2 ⟹ 𝑎𝐺𝑥 = 𝑔𝑠𝑖𝑛𝜃 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.51 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion HCM City Univ of Technology, Mechanical Engineering Department System Dynamics Modeling of Rigid Body Mechanical Systems §4.General Planar Motion b Moment equation about the the point 𝑃 𝑀𝑃 = 𝐼𝐺 𝛼 + 𝑚𝑎𝐺𝑥 𝑑 𝑑 = 𝑟, 𝑀𝑃 = 𝑚𝑔𝑠𝑖𝑛𝜃 𝑟 𝑎𝐺 ⟹ 𝑚𝑔𝑟𝑠𝑖𝑛𝜃 = 𝐼𝐺 𝑥 + 𝑚𝑎𝐺𝑥 𝑟 𝑟 𝑚𝑔𝑟 𝑠𝑖𝑛𝜃 ⟹ 𝑎𝐺𝑥 = 𝑚𝑟 + 𝐼𝐺 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.52 Nguyen Tan Tien 2.53 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion - Example 2.4.3 Maximum Vehicle Acceleration Determine the maximum acceleration of the rear-wheel drive vehicle as a function of the friction coefficient 𝜇𝑠 The vehicle mass is 1800𝑘𝑔, and its dimensions are 𝐿𝐴 = 1.3𝑚, 𝐿𝐵 = 1𝑚, and 𝐻 = 0.5𝑚 c The maximum possible force 𝐹𝑚𝑎𝑥 = 𝜇𝑠 𝑚𝑔𝑐𝑜𝑠𝜃 𝐼𝐺 𝛼 𝐼𝐺 𝑎𝐺𝑥 𝐼𝐺 𝑚𝑔𝑠𝑖𝑛𝜃 𝐹= = = 𝑟 𝑟 𝐼𝐺 + 𝑚𝑟 If 𝐹𝑚𝑎𝑥 > 𝐹, the cylinder will not slip The condition of no slip 𝐼𝐺 𝑠𝑖𝑛𝜃 𝜇𝑠 𝑐𝑜𝑠𝜃 > 𝐼𝐺 + 𝑚𝑟 For a solid cylinder, this reduces to 𝜇𝑠 𝑐𝑜𝑠𝜃 > 𝑠𝑖𝑛𝜃 HCM City Univ of Technology, Mechanical Engineering Department System Dynamics §4.General Planar Motion Solution 2.54 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems Maximum traction force (maximum acceleration) is obtained when the driving tires are just on the verge of slipping relative to the road surface Apply Newton’s law 𝑓𝑥 = 𝑚𝑎𝐺𝑥 ⟹ 𝜇𝑠 𝑁𝐵 = 𝑚𝑎𝐺𝑥 𝑓𝑦 = 𝑚𝑎𝐺𝑦 ⟹ 𝑁𝐴 + 𝑁𝐵 − 𝑚𝑔 = The moment equation about the mass center 𝐺 𝑀𝐺 = 𝐼𝐺 𝛼 ⟹ 𝑁𝐵 𝐿𝐵 − 𝜇𝑠 𝑁𝐵 𝐻 − 𝑁𝐴 𝐿𝐴 = 𝑚𝑔𝐿𝐴 9.8 × 1.3𝑚 25.5 ⟹ 𝑁𝐵 = = = 𝑚 𝐿𝐴 + 𝐿𝐵 − 𝜇𝑠 𝐻 1.3 + − 0.5𝜇𝑠 4.6 − 𝜇𝑠 The maximum acceleration 𝜇𝑠 𝑁𝐵 25.5𝜇𝑠 𝑎𝐺𝑥 = = 𝑚 4.6 − 𝜇𝑠 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 8/25/2013 System Dynamics 2.55 Modeling of Rigid Body Mechanical Systems §4.General Planar Motion 4.Dymamics of a Personal Transporter - The transporter’s motors drive the wheels to balance the vehicle with the help of a computer-control system using tilt sensors and gyroscopes - The dynamics of a personal transporter are similar to a classic control problem called the inverted pendulum System Dynamics 2.56 §4.General Planar Motion - Example 2.4.4 Modeling of Rigid Body Mechanical Systems Transporter Equations of Motion Model the transporter as a cart of mass 𝑀 and an inverted pendulum attached to the cart by a pivot at point 𝑃 The pendulum mass is 𝑚 and its center of mass 𝐺 is a distance 𝐿 from 𝑃 The inertia of the pendulum about 𝐺 is 𝐼𝐺 For generality we include an applied torque 𝑇 about the pivot, which is due to a motor at the pivot in some applications Derive the equations of motion with 𝑓 and 𝑇 as the inputs, and 𝑥 and 𝜙 as the outputs HCM City Univ of Technology, Mechanical Engineering Department System Dynamics 2.57 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion Solution System Dynamics 2.58 Nguyen Tan Tien Modeling of Rigid Body Mechanical Systems §4.General Planar Motion Consider the pendulum 𝑑2 𝑚 𝑥 − 𝐿𝑠𝑖𝑛𝜙 = 𝐻 𝑑𝑡 ⟹ 𝑚𝑥 − 𝑚𝐿(−𝑠𝑖𝑛𝜙𝜙2 + 𝑐𝑜𝑠𝜙𝜙) = 𝐻 The moment equation about the pendulum’s pivot point 𝑃 𝐼𝐺 + 𝑚𝐿2 𝜙 − 𝑚𝐿𝑥𝑐𝑜𝑠𝜙 = 𝑇 + 𝑚𝑔𝐿𝑠𝑖𝑛𝜙 HCM City Univ of Technology, Mechanical Engineering Department HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien Consider the base 𝑀𝑥 = −𝐻 − 𝑓 ⟹ 𝐻 = −𝑀𝑥 − 𝑓 The equation of motion of inverted pendulum 𝑚 + 𝑀 𝑥 − 𝑚𝐿 𝑐𝑜𝑠𝜙𝜙 − 𝑠𝑖𝑛𝜙𝜙2 = −𝑓 HCM City Univ of Technology, Mechanical Engineering Department Nguyen Tan Tien 10 ...