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]
