Answer:
a) [tex]\theta = 52\,rad[/tex], b) [tex]v_{A} = 44\cdot r \,\frac{m}{s}[/tex], c) [tex]a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}[/tex], [tex]a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}[/tex]
Explanation:
a) The angular motion is obtained by integrating the angular acceleration function twice:
[tex]\alpha = 3\cdot t^{2} + 12[/tex]
[tex]\omega = t^{3} + 12\cdot t + 12[/tex]
[tex]\theta = \frac{1}{4}\cdot t^{4} + 6\cdot t^{2} + 12\cdot t[/tex]
The angular motion when t = 2 s. is:
[tex]\theta = \frac{1}{4}\cdot (2\,s)^{4} + 6\cdot (2\,s)^{2} + 12\cdot (2\,s)[/tex]
[tex]\theta = 52\,rad[/tex]
b) Let be [tex]r[/tex] the distance between A and the rotation axis, measured in meters. The magnitude of the angular velocity when t = 2 s. is:
[tex]\omega = (2\,s)^{3} + 12\cdot (2\,s) + 12[/tex]
[tex]\omega = 44\,\frac{rad}{s}[/tex]
Finally, the magnitude of the velocity is:
[tex]v_{A} = 44\cdot r \,\frac{m}{s}[/tex]
c) The angular acceleration of the disk when t = 2 s. is:
[tex]\alpha = 3\cdot (2\,s)^{2} + 12[/tex]
[tex]\alpha = 24\,\frac{rad}{s^{2}}[/tex]
Lastly, the normal and tangential components at point A are, respectively:
[tex]a_{A,n} = \omega^{2}\cdot r[/tex]
[tex]a_{A,n} = \left(44\right)^{2}\cdot r[/tex]
[tex]a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}[/tex]
[tex]a_{A,t} = \alpha \cdot r[/tex]
[tex]a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}[/tex]
