Respuesta :
Answer:
a)[tex]F=\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)[/tex]
b)[tex]V=\omega \sqrt{\dfrac{L^2-r^2}{2}}[/tex]
c)[tex]t=\dfrac{\pi}{\omega \sqrt2}[/tex]
Explanation:
Given that
Length =L
Mass = m
Force on elemental part
dF= dm ω² r
dm = m/L dr
dF= ω² r m/L dr
By integrating from r to L
[tex]F=\dfrac{m}{L} \int_{r}^{L}\omega^2 r \ dr[/tex]
[tex]F=\dfrac{\omega^2 m}{L}\left(\dfrac{L^2}{2}-\dfrac{r^2}{2}\right)[/tex]
[tex]F=\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)[/tex]
Velocity V
[tex]V=\sqrt{\dfrac{F}{\dfrac{m}{L}}}[/tex]
[tex]V=\sqrt{\dfrac{\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)}{\dfrac{m}{L}}}[/tex]
[tex]V=\omega \sqrt{\dfrac{L^2-r^2}{2}}[/tex]
[tex]\dfrac{dr}{dt}=V=\omega \sqrt{\dfrac{L^2-r^2}{2}}[/tex]
[tex]\int_{0}^{L}\dfrac{dr}{\sqrt{L^2-r^2}}=\int_{0}^{t} \dfrac{\omega }{\sqrt2}dt[/tex]
[tex]sin^{-1}1=\ \dfrac{\omega }{\sqrt2}t[/tex]
[tex]\dfrac{\pi}{2}=\ \dfrac{\omega }{\sqrt2}t[/tex]
[tex]t=\dfrac{\pi}{\omega \sqrt2}[/tex]
