Answer:
a).[tex]x_{t}=583.3rad[/tex]
b).[tex]t_{total}=10.52s[/tex]
c).[tex]a=12.62 \frac{rad}{s^2}[/tex]
Explanation:
The angular acceleration is constant so we can use the formulas of uniform motion with the model of angular acceleration
a).
[tex]x_{r}=x_{i}+v_{i}+\frac{1}{2}a_{a}*t^2[/tex]
[tex]x_{r}=0+28.0\frac{rad}{s}*2.20s+\frac{1}{2}*35.0\frac{rad}{s^2}*2.20s[/tex]
[tex]x_{r}=146.3rad[/tex]
so the total angle between t=0 and the time it stopped is
[tex]x_{t}=146.3rad+437rad=583.3rad[/tex]
b).
[tex]w_{f}=w_{o}+a*t[/tex]
[tex]w_{f}=28.0rad/s+35rad/s^2*2.2s=105rad/s[/tex]=[tex]w_{o}[/tex]
[tex]x_{t}-x_{r}=\frac{1}{2}*(w_{o}-w_{f})*t[/tex]
[tex]583.3-146.3=\frac{1}{2}*(105-0)*t[/tex]
[tex]t=\frac{437rad}{105\frac{rad}{s}}=8.32s[/tex]
[tex]t_{total}=8.32+2.2=10.52s[/tex]
c).
[tex]w_{f}=w_{o}+a*t[/tex]
[tex]0=105 rad/s+a*8.32s[/tex]
[tex]a=\frac{105 rad/s}{8.32s}[/tex]
[tex]a=12.62 \frac{rad}{s^2}[/tex]
