RotationalMechanicalSystem

RotationalMechanicalSystem - ME375 Handouts Rotational...

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Unformatted text preview: ME375 Handouts Rotational Mechanical Systems • • • • • Variables Basic Modeling Elements Interconnection Laws Derive Equation of Motion (EOM) Energy Transfer School of Mechanical Engineering Purdue University ME375 Rotation - 1 1 ME375 Handouts Variables • θ : angular displacement [rad] [rad] • ω : angular velocity [rad/sec] [rad/sec] • α : acceleration [rad/sec2] [rad/sec • τ : torque [Nm] • p : power [Nm/sec] [Nm/sec] • w : work ( energy ) [Nm] [Nm] 1 [Nm] = 1 [J] (Joule) d & θ =θ =ω dt d dd d2 && & ω =ω = θ = 2θ =θ =α dt dt dt dt &d p = τ ⋅ω = τ ⋅θ = w dt FI HK w (t1 ) = w (t0 ) + = w ( t0 ) + School of Mechanical Engineering Purdue University z z t1 t0 t1 t0 p(t ) dt & (τ ⋅ θ ) dt ME375 Rotation - 2 2 ME375 Handouts Basic Rotational Modeling Elements • Spring • Damper – Stiffness Element τS – Friction Element θ2 θ1 τS b τ S = K θ 2 − θ1 g d ib & & τ D = B θ2 − θ1 = B ω 2 − ω1 – Analogous to Translational Spring. – Stores Potential Energy. – E.g., shafts g – Analogous to Translational Damper. – Dissipates Energy. – E.g., bearings, bushings, ... School of Mechanical Engineering Purdue University ME375 Rotation - 3 3 ME375 Handouts Basic Rotational Modeling Elements • Moment of Inertia of Inertia • Parallel Axis Theorem Axis Theorem J = JO + M r 2 – Inertia Element Ex: d J = JO + M r 2 = && Jθ = ∑τ i i J = JO + M r 2 – Analogous to Mass in Translational Motion. – Stores Kinetic Energy. = School of Mechanical Engineering Purdue University ME375 Rotation - 4 4 ME375 Handouts Interconnection Laws • Newton’s Second Law af d && J ω = J θ = ∑ τ EXTi 2 dt 1 3 i Angular Momentum • Newton’s Third Law – Action & Reaction Torque • Angular Displacement Law School of Mechanical Engineering Purdue University ME375 Rotation - 5 5 ME375 Handouts Example Deri Derive a model (EOM) for the model (EOM) for the following following system: τ (t) J θ K B FBD: School of Mechanical Engineering Purdue University ME375 Rotation - 6 6 ME375 Handouts Energy Distribution • EOM of a simple Mass-Spring-Damper System of simple Mass System && & J θ + Bθ + K θ = τ ( t ) 0 Contribution of Inertia 1 1 Contribution of the Damper Contribution of the Spring 3 Total Applied Torque We want to look at the energy distribution of the system. How should we start ? • Multiply the above equation by angular velocity term ω : ⇐ What have we done ? bg && & && & J θ ⋅ θ + Bθ ⋅ θ + K θ ⋅ θ = τ t ⋅ ω • ⇐ What are we doing now ? Integrate the second equation w.r.t. time: z t1 && & J θ ⋅ θ dt t0 142 4 3 ΔKE = 1 J ω 2 2 E + z t1 B ω ⋅ ω dt t0 1 42 43 z t1 2 t0 B ω dt E ≥0 + z t1 & K θ ⋅ θ dt t0 1 42 43 Δ PE = 1 K θ 2 2 E School of Mechanical Engineering Purdue University = z t1 bg τ t ⋅ ω dt t0 1 42 43 ΔE T ota l w or k d one b y t h e applied torque τ ( t ) from tim e t 0 to t1 ME375 Rotation - 7 7 ME375 Handouts Gears N2 N1 R2 R1 A ω2 ω1 P θ1 θ2 B A B School of Mechanical Engineering Purdue University ME375 Rotation - 8 8 ME375 Handouts Gear Train θ1 N1 J1 EOM: θ2 T J2 N2 FBD: J1 T R1 θ1 F F R2 θ2 J2 School of Mechanical Engineering Purdue University ME375 Rotation - 9 9 ME375 Handouts Example • Rolling without slipping without slipping Coefficient of friction μ x θ R K B f(t) J, M FBD: School of Mechanical Engineering Purdue University ME375 Rotation - 10 10 ME375 Handouts Example (cont.) (cont.) School of Mechanical Engineering Purdue University ME375 Rotation - 11 11 ...
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