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

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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  1. Answer:

    [tex]g = \frac{V_B^2}{L} = \frac{\Delta y}{\Delta x}[/tex]

    Explanation:

    Given that :

    length of the thin rod = L

    mass = m

    The rotational inertia I = [tex]\frac{ML^2}{3}[/tex]

    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

    [tex]I =\int\limits \ r^2 \, dm[/tex]

    where :

    [tex]\lambda = \frac{m}{L} \\ \\ m = L \lambda \\\\ \lambda dr = dm[/tex]

    [tex]I =\int\limits^L_0 { \lambda r^2 } \, dr[/tex]

    [tex]I = \frac{\lambda r^3}{3} |^L___0[/tex]

    [tex]I = \frac{m}{3 L}L^3[/tex]

    [tex]I = \frac{1}{3 }mL^2[/tex]

    [tex]k_f = \mu___0}}} = \frac{1}2} I \omega^2[/tex]

    where:

    [tex]V_B = \omega L[/tex]

    [tex]\omega = \frac{V_B}{L}[/tex]

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

    [tex]\frac{1}{2} (\frac{1}{3}mL^2)\frac{V_B^2}{L^2} = mg \frac{L}{2}[/tex]

    [tex]V_B^2 = 3gL[/tex]     since m = 3g

    where :

    [tex]V_B^2 =[/tex]vertical axis on the graph

    L = horizontal axis

    [tex]V_B^2 = 3gL[/tex]     ( y = mx)

    [tex]3g = \frac{V_B^2}{L}[/tex]

    [tex]g = \frac{V_B^2}{L} = \frac{\Delta y}{\Delta x}[/tex]

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