Theory of Machines Miscellaneous

Theory of Machines Miscellaneous

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Theory of Machines

Theory of Machines Miscellaneous

Direction: A uniform rigid slender bar of mass 10 kg, hinged at the left end is suspended with the help of spring and damper arrangement as shown in the figure where K= 2 kN/m, C = 500 Ns/m and the stiffness of the torsional spring kθ is 1 kN /m/rad. Ignore the hinge dimensions.

  1. The undamped natural frequency of oscillations of the bar about the hinge point is








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    x1 = l1Q
    x2 = Q.l2
    For small deflection, after equilibrium

    Taking momentum about O

    C
    dx1
    		. l1 + kx2l2 + Iα + kθ.Q = 0 ...(i)
    dt

    Correct Option: A

    x1 = l1Q
    x2 = Q.l2
    For small deflection, after equilibrium

    Taking momentum about O

    C
    dx1
    		. l1 + kx2l2 + Iα + kθ.Q = 0 ...(i)
    dt

  1. If Cf is the coefficient of speed fluctuation of a flywheel then the ratio of ωmax /ωmin will be








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    Cf =
    ωmax - ωmin
    ωmax + ωmin
    2

    ⇒
    ωmax
    		=
    2 + Cf
    ωmin2 - Cf

    Correct Option: D

    Cf =
    ωmax - ωmin
    ωmax + ωmin
    2

    ⇒
    ωmax
    		=
    2 + Cf
    ωmin2 - Cf

  1. Which of the following statements is correct?








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    Flywheel reduces speed fluctuations during a cycle for a constant load, but flywheel does not control the mean speed of the engine if the load changes.

    Correct Option: A

    Flywheel reduces speed fluctuations during a cycle for a constant load, but flywheel does not control the mean speed of the engine if the load changes.

  1. A flywheel of moment of inertia 9.8 kg m2 fluctuates by 30 rpm for a fluctuation in energy of 1936 Joules. The mean speed of the flywheel is (in rpm)








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    ∆E = mR2ω2∆S
    = mR2ω(ω1 – ω2)

    ∴ 1936 = 9.8 × ω
    2π
    		× 30
    60

    or Mean speed of flywheel, ω= 600 rpm

    Correct Option: A

    ∆E = mR2ω2∆S
    = mR2ω(ω1 – ω2)

    ∴ 1936 = 9.8 × ω
    2π
    		× 30
    60

    or Mean speed of flywheel, ω= 600 rpm

  1. An epicyclic gear train is shown schematically in the adjacent figure. The sun gear 2 on the input shaft is a 20 teeth external gear. The planet gear 3 is a 40 teeth external gear. The ring gear 5 is a 100 teeth internal gear. The ring gear 5 is fixed and the gear 2 is rotating at 60 rpm ccw(ccw-counterclockwise and cw- clockwise)

    The arm 4 attached to the output shaft will rotate at








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    T2 = 20; T3 = 40; T5 =100

    Given y –x
    T2
    		= 0
    T5

    ⇒ x = 5y ....(1)
    and y + x = 60 ....(2)
    Solving (1) and (2), we get
    y + 5y = 60
    ⇒ y = 10 rpm ccw

    Correct Option: B

    T2 = 20; T3 = 40; T5 =100

    Given y –x
    T2
    		= 0
    T5

    ⇒ x = 5y ....(1)
    and y + x = 60 ....(2)
    Solving (1) and (2), we get
    y + 5y = 60
    ⇒ y = 10 rpm ccw