Respuesta :
Explanation:
Angular acceleration of the bar, [tex]\alpha=(13+7t)\ rad/s^2[/tex]..........(1)
We need to find the angle in radians through which the bar turns in the first 4.69 s.
We know that,
[tex]\alpha=\dfrac{d\omega}{dt}[/tex]
[tex]\omega[/tex] is angular velocity
On integrating equation (1) wrt t
[tex]\omega=\int\limits {\alpha .dt }[/tex]
[tex]\omega=13t+\dfrac{7t^2}{2}[/tex]...............(2)
Also, [tex]\omega=\dfrac{d\theta}{dt}[/tex]
On integrating equation (2) wrt t as :
[tex]\theta=\int\limits{\omega.dt}[/tex]
[tex]\theta=\int\limits^{4.69}_0 {({13t+\dfrac{7t^2}{2}}).dt}[/tex]
[tex]\theta=\dfrac{13t^2}{2}+\dfrac{7t^3}{6}[/tex]
At t = 4.69 s
[tex]\theta=\dfrac{13(4.69)^2}{2}+\dfrac{7(4.69)^3}{6}[/tex]
[tex]\theta=263.32\ radians[/tex]
Hence, this is the required solution.
