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EGR 280 Mechanics 14 – Kinematics of Rigid Bodies.

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Presentation on theme: "EGR 280 Mechanics 14 – Kinematics of Rigid Bodies."— Presentation transcript:

1 EGR 280 Mechanics 14 – Kinematics of Rigid Bodies

2 Kinematics of rigid bodies We will group the various types of rigid-body motion into three categories: Translation – rectilinear or curvilinear motion where all of the particles that make up the body move along parallel paths Rotation about a fixed axis – the particles of the rigid body move in parallel planes along circles centered on the axis of rotation General plane motion – plane motion that is a combination of translation and rotation

3 Translation Consider a rigid body in translation. Let A and B be points on that body. r B = r A + r B/A Note that r B/A does not change magnitude or direction. Therefore v A = v B and a A = a B If a rigid body is in translation, all of the points of the body have the same velocity and the same acceleration at any given instant. X Y Z rArA A B rBrB r B/A

4 Rotation about a fixed axis Consider a rigid body rotating about a fixed axis. Let P be a point of the body and r be its position with respect to a fixed point O. Angle θ is the angular coordinate. The velocity of P is tangent to the circle that P makes around the axis of rotation. Graphics and problem statements © 2004 R.C. Hibbeler. Published by Pearson Education, Inc., Upper Saddle River, NJ. Let ω = dθ/dt be the angular velocity of the body, acting in the direction (by the RHR) of the axis of rotation. The velocity of P can be written as v = dr/dt = ω × r

5 and the acceleration is, by definition a = dv/dt = d(ω × r)/dt = (dω/dt × r) + (ω × dr/dt) = α × r + ω × (ω × r) where α = dω/dt is the angular acceleration. Two-dimensional fixed-point rotation If the motion takes place in two dimensions, then the angular velocity and angular acceleration will always be perpendicular to the plane of the motion. v = ω × r = ωk × r (v = rω) a = α × r + ω × (ω × r) = αk × r + ωk × (ωk × r) = αk × r – ω 2 r X Y r ωkωk v

6 Note that the acceleration has two components, one directed toward O with magnitude equal to rω 2, and the other tangent to the circular path of P with magnitude equal α (compare to a = (dv/dt)e t + (v 2 /ρ)e n ). We have:ω = dθ/dt α = dω/dt = d 2 θ/dt 2 α = dω/dt = (dω/dθ)(dθ/dt) = ω(dω/dθ) Special cases: Uniform rotation: α = 0 θ = θ 0 + ωt Uniformly accelerated rotation: α constant θ = θ 0 + ω 0 t + ½ αt 2 ω = ω 0 + αt ω 2 = ω 0 2 + 2αθ

7 General plane motion Any plane motion can be considered as the sum of a translation and a rotation: v B = v A + v B/A where v B/A is simply a rotation about A: v B/A = ωk × r B/A v B/A = ωr B/A v B = v A + ωk × r B/A Velocity is a property of a point; angular velocity is a property of a body The sense of the relative velocity may change with a change in reference point X Y Z rArA A B rBrB r B/A

8 and the absolute acceleration of any point B on the rigid body can be written as: a B = a A + a B/A = a A + α × r B/A + ω × (ω × r B/A ) = a A + αk × r B/A + ωk × (ωk × r B/A ) = a A + αk × r B/A – ω 2 r B/A

9 Special cases: Rolling If a circular cylinder rolls without slipping on a surface: v B = v A + ωk × r B/A 0 = -vi + ωk × (-rj) v = ωr If two pulleys are joined by a belt: v B = v A + ω 1 k × r B/A = ω 1 k × r B/A v C = v D + ω 2 k × r C/D = ω 2 k × r C/D If the belt doesn’t slip or stretch, v B =v C and ω 1 r 1 = ω 2 r 2 B ω v r A B A C D ω1ω1 ω2ω2

10 Special cases: Rolling When the teeth of two gears mesh, their pitch circles roll on each other without slipping: The velocity of point C on gear A must be the same as the velocity of point C on gear B: v C = v A + ω 1 k × r C/A = -ω 1 k × r C/A v C = v B + ω 2 k × r C/B = ω 2 k × r C/B ω 1 r 1 = -ω 2 r 2 The number of teeth on each gear, N, must be an integer value, and is proportional to the pitch diameter: ω 1 N 1 = -ω 2 N 2 A B C ω2ω2 ω1ω1 vCvC


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