Answer:
The value of angular acceleration after 3.41 s is -0.18925 rad/s^2
Step-by-step explanation:
As per the given data the value is given as
[tex]\frac{d\theta}{dt}=\omega_oe^{-\sigma t}[/tex]
Where
[tex]3.63=4.43 e^{-\sigma *3.92}\\e^{-\sigma 3.92}=\frac{363}{443}\\-\sigma \cdot \:3.92\ln \left(e\right)=\ln \left(\frac{363}{443}\right)\\\sigma=0.05080[/tex]
Now as the acceleration is given as
[tex]\alpha=\frac{d}{dt}(\frac{d\theta}{dt})\\\alpha=\frac{d}{dt}(\omega_oe^{-\sigma t})\\\alpha=\omega_o\frac{d}{dt}(e^{-\sigma t})\\\alpha=-\omega_o \sigma(e^{-\sigma t})[/tex]
As values of
thus acceleration is given as
[tex]\alpha=-\omega_o \sigma(e^{-\sigma t})\\\alpha=-4.43*0.05080(e^{-0.05080* 3.41})\\\alpha=-0.18925\, rad/s^2[/tex]
So the value of angular acceleration after 3.41 s is -0.18925 rad/s^2
