A uniform, thin rod of length L and mass M is allowed to pivot about its end. The rotational inertia of a rod about its end is ML2/3. The ro

Question

A uniform, thin rod of length L and mass M is allowed to pivot about its end. The rotational inertia of a rod about its end is ML2/3. The rod is fixed at one end and allowed to fall from the horizontal position through the vertical position. The student has a variety of rods of various lengths that all have a uniform mass distribution. Design an experiment that the student could conduct to find an experimental value of g.

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Doris 6 months 2021-07-17T15:41:47+00:00 1 Answers 21 views 0

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    2021-07-17T15:43:14+00:00

    Answer:

    g = \frac{V_B^2}{L}  =  \frac{\Delta y}{\Delta x}

    Explanation:

    Given that :

    length of the thin rod = L

    mass = m

    The rotational inertia I = \frac{ML^2}{3}

    The experimental design that the student can use to conduct the experimental value of g can be determined as follow:

    Taking the integral value of I

    I =\int\limits \  r^2  \, dm

    where :

    \lambda = \frac{m}{L} \\ \\ m = L \lambda \\\\ \lambda dr = dm

    I =\int\limits^L_0 { \lambda r^2 } \, dr

    I = \frac{\lambda r^3}{3} |^L___0

    I = \frac{m}{3 L}L^3

    I = \frac{1}{3 }mL^2

    k_f = \mu___0}}} = \frac{1}2} I \omega^2

    where:

    V_B = \omega L

    \omega = \frac{V_B}{L}

    Equating: \frac{1}2} I \omega^2 = mg \frac{L}{2}; we have:

    \frac{1}{2} (\frac{1}{3}mL^2)\frac{V_B^2}{L^2} = mg \frac{L}{2}

    V_B^2 = 3gL     since m = 3g

    where :

    V_B^2 =vertical axis on the graph

    L = horizontal axis

    V_B^2 = 3gL     ( y = mx)

    3g = \frac{V_B^2}{L}

    g = \frac{V_B^2}{L}  =  \frac{\Delta y}{\Delta x}

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