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    Rigid Body Dynamics

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    (Please refer attachment for detailed problem description)

    Problem 1 : The 40 kg disc is released from rest with the spring compressed. At the instant shown (see attachment) it has a speed of 4 m/s and the spring is unstretched. From this poinr determine the distance d that the disc moves down the 30 degree ramp before momentarily stopping. The disc rolls without slipping. Determine :

    a) The distance d in meters down the ramp from point A.
    b) The moment of inertia of the disc about G in kgm^2
    c) The angular velocity of the disc in rad/sec at this instant.
    d) The work done by the spring in Nm during the downward motion.
    e) The distance h in m, up the ramp from A to the starting point.

    Problem 2 : A 5 kg uniform rod 800mm long rotates about pin B in a vertical plane. If the rod is released from rest when it is horizontal, determine :

    a) Moment of inertia of the rod about B in kg m^2.
    b) Angular velocity of the rod in rad/sec when vertical.
    c) Angular velocity of the rod in kg m^2 at theta = 75 degrees
    d) X direction support reaction in N at pin B at theta = 75 degrees
    e) Y direction support reaction in N at pin B at theta = 75 degrees

    © BrainMass Inc. brainmass.com October 9, 2019, 9:11 pm ad1c9bdddf
    https://brainmass.com/math/discrete-math/rigid-body-dynamics-175831

    Attachments

    Solution Preview

    Please refer attachment.

    Problem 1

    k = 200 N/m
    4 m/s
    G Mass = 40 kg, Radius = 0.3m

    A d
    dsin30O

    30O

    Moment of inertia of the disc about an axis through centre of the disc = I = ½ MR2 = ½ x 40 x 0.32 = 1.8 kgm2

    At the given moment the disc has a translational velocity v = 4 m/s.

    Its angular velocity ω = v/R = 4/0.3 = 13.33 radians/sec

    At the given moment the disc has a translational kinetic energy = ½ mv2 = ½ x 40 x 42 = 320 J

    And rotational KE = ½ Iω2 = ½ x 1.8 x (13.33)2 = 160 J

    Total KE = 320 + 160 = 480 J

    With reference to the centre of the disc at the lowest position, the height of the disc in the given position is dsin30O = 0.5d.

    Potential energy of the disc in the given position with reference to its lowest position = mgh = 40x9.81x0.5d = 196d

    Total mechanical energy of the disc in the given position = KE +PE ...

    Solution Summary

    A rigid body dynamics is analyzed. The moment of inertia of the disc about G in kgm^2 is analyzed.

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