Theory of Machines Miscellaneous


Theory of Machines Miscellaneous

  1. A flywheel connected to a punching machine has to supply energy of 400 Nm while running at a mean angular speed of 20 rad/s. If the total fluctuation of speed is not to exceed ± 2% the mass moment of inertia of the flywheel in kg-m² is









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    ΔE = 400Nm
    ωmean = 20 rad/s
    CS = ± 2%

    = 2 ×
    2
    = 0.04
    100

    ΔE = Iω²meanCS
    400 = I(20)² × 0.04
    I = 25 kg-m²

    Correct Option: A

    ΔE = 400Nm
    ωmean = 20 rad/s
    CS = ± 2%

    = 2 ×
    2
    = 0.04
    100

    ΔE = Iω²meanCS
    400 = I(20)² × 0.04
    I = 25 kg-m²


  1. A gear train is made up of five spur gears as shown in the figure. Gear 2 is driver and gear 6 is driven member. N2, N3, N4, N5 and N7 represent number of teeth on gears 2,3,4,5 and 6 respectively. The gear(s) which act(s) as idler(s) is/are









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    ω2
    =
    ω2
    ω3
    ω5
    =
    N3
    N5
    N6
    =
    N3N6
    ω6ω3ω5ω6N2N4N5N3N4

    only Gear(5) is Idle

    Correct Option: C

    ω2
    =
    ω2
    ω3
    ω5
    =
    N3
    N5
    N6
    =
    N3N6
    ω6ω3ω5ω6N2N4N5N3N4

    only Gear(5) is Idle



  1. For the epicyclic gear arrangement shown in the figure, ω2 = 100 rad/s clockwise (cw) and ωarm = 80 rad/s counter clockwise (ccw). The angular velocity ω5 (in rad/s) is









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    Given: ωarm = – 80 rad/sec = Y (CCW)
    ω2 = 100 rad/sec = X + Y (CW)
    ∴ X = 100 – Y = 100 + 8 = 108 rad/sec (CW)

    Also, ωs = -
    1
    X + Y = -
    1
    (180) + (-80)
    33

    = – 140 rad/sec (CCW)
    ∴ ωs = 140 rad/sec (CW)
    Note: CCW → Counter Clock Wise
    CW → Clock Wise

    Correct Option: B


    Given: ωarm = – 80 rad/sec = Y (CCW)
    ω2 = 100 rad/sec = X + Y (CW)
    ∴ X = 100 – Y = 100 + 8 = 108 rad/sec (CW)

    Also, ωs = -
    1
    X + Y = -
    1
    (180) + (-80)
    33

    = – 140 rad/sec (CCW)
    ∴ ωs = 140 rad/sec (CW)
    Note: CCW → Counter Clock Wise
    CW → Clock Wise


  1. In the figure, link 2 rotates with constant angular velocity ω2. A slider link 3 moves outwards with a constant relative velocity VQ/P, where Q is a point on slider 3 and P is a point on link 2. The magnitude and direction of Coriolis component of acceleration is given by









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    Direction is obtained by rotating velocity vector through 90º in the direction of rotation of the link.

    Correct Option: A

    Direction is obtained by rotating velocity vector through 90º in the direction of rotation of the link.



  1. The number of degrees of freedom of a five link plane mechanism with five revolute pairs as shown in the figure is









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    DOF = 3(n – 1) – 2j – h n = 5, j = 5
    DOF = 3(5 – 1) – 2 × 5 = 2

    Correct Option: C

    DOF = 3(n – 1) – 2j – h n = 5, j = 5
    DOF = 3(5 – 1) – 2 × 5 = 2