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.

Answer 1

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)

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

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