When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequency, ωω. Assum

Question

When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequency, ωω. Assume that this frequency is a function of the diameter, DD, the mass of the cylinder, mm, and the specific weight γγ, of the liquid. Determine, with the aid of dimensional analysis, how the frequency is related to these variables. If the mass of the cylinder were increased, would the frequency increase or decrease?

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Linh Đan 2 months 2021-07-19T14:33:21+00:00 1 Answers 3 views 0

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    2021-07-19T14:34:41+00:00

    Answer:

    w = √[g /L (½ r²/L2 + 2/3 ) ]

    When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

    Explanation:

    We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

              w² = mg d / I

    In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

             d = L / 2

    The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

            I = ¼ m r2 + ⅓ m L2

            I = m (¼ r2 + ⅓ L2)

    now let’s use the concept of density to calculate the mass of the system

            ρ = m / V

            m = ρ V

    the volume of a cylinder is

             V = π r² L

              m =  ρ π r² L

    let’s substitute

            w² = m g (L / 2) / m (¼ r² + ⅓ L²)

            w² = g L / (½ r² + 2/3 L²)

            L >> r

             w = √[g /L (½ r²/L2 + 2/3 ) ]

    When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

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