 Ch 5. Rigid Body General Motion Multimedia Engineering Dynamics Fixed Axis Rotation Plane Motion Velocities Zero Velocity Point Plane Motion Accelerations Multiple Gears Rot. Coord. Velocities Rot. Coord. Acceleration
 Chapter - Particle - 1. General Motion 2. Force & Accel. 3. Energy 4. Momentum - Rigid Body - 5. General Motion 6. Force & Accel. 7. Energy 8. Momentum 9. 3-D Motion 10. Vibrations Appendix Basic Math Units Basic Equations Sections Search eBooks Dynamics Statics Mechanics Fluids Thermodynamics Math Author(s): Kurt Gramoll ©Kurt Gramoll DYNAMICS - EXAMPLE Two Connected Bars Example Two rods are pinned to a fixed point at one end, and connect to each other at the other end using a slider joint. Point C can freely move along rod DE but is pinned to rod AB at C. Rod AB rotates clockwise with angular velocity ωAB = 4 rad/s and angular acceleration αAB = 5 rad/s2 when θ = 45o. Find the angular acceleration of rod DE at this instant. The collar at C is pin connected to AB and slides over rod DE. Solution Since the distance between point D and C varies, two frames of reference is useful to solve the problem. Attach the local x-y coordinate to the rotating AE rod, and the global X-Y coordinate to the non-moving background. Velocities: Before accelerations can be found, the velocity of point C, vC/D, and angular velocity of ΩDE (= ΩDC) needs to be determined.      vC = vD + Ω × rC/D + ( vC/D)rel      vC = vD + ΩDC× rC/D + vC/D      vD = 0      ωAB = 4k      rC/D = 0.5 m i      ΩDC = ωDC Since distance from A and C does not change, velocity of C can be determined from bar AB      vC = ωAB × rC/A = -4k × (0.5i + 0.5j)           = 2i - 2j m/s Putting all terms together gives,      2i - 2j = 0 + (-ωDC k) × (0.5i) + vC/D reli      2i - 2j = -0.5ωDC j + vC/D rel i Comparing the i and j terms give      vC/D = 2 m/s   and   ωDC = 4 rad/s Accelerations:      aC = aD + dΩDC/dt × rC/D + ΩDC× (ΩDC × rC/D)               + 2ΩDC × vC/D + aC/D      aC = aD + αDE × rC/D + ωDE× ( ωDE × rC/D)               + 2ωDE × vC/D + aC/D where,      aD = 0      rC/D = 0.5i m      αDC = (-αDCk) × 0.5i = -0.5αDEj m/s2      ωDC × ( ωDC × rC/D) = (- 4)2 (0.5 i) = -8i rad/s      2ωDC × vC/D = 2 (-4k) × 2i = 16j m/s2      aC/D = aC/Di Acceleration of C can be determined from bar AC      aC = αAC × rC/A - ω2 rC/A          = -5k × (0.5i + 0.5j) - (4)2 (0.5i + 0.5j)          = -5.5i - 10.5j m/s2 -5.5i - 10.5j = 0 + (-αDEk)× 0.5i - (- 4)2 (0.5 i)        + 2 (-4k)×2i + (aC/D)i Putting all terms together gives,      -5.5i - 10.5j = 0 - 0.5αDCj - 8i - 16j + aC/Di Comparing the j terms      -10.5 = -16 - 0.5αDE      αDC = -11 rad/s2 αDC = 11k rad/s2 (counter-clockwise)

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