Answer:
a) [tex]\omega = 50\,\frac{rad}{s}[/tex], b) [tex]\omega = 0\,\frac{rad}{s}[/tex]
Explanation:
The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:
[tex]\tau = F \cdot r[/tex]
[tex]\tau = m\cdot a \cdot r[/tex]
[tex]\tau = m \cdot \alpha \cdot r^{2}[/tex]
Where [tex]\alpha[/tex] is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:
[tex]\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}[/tex]
[tex]\alpha = -3\,\frac{rad}{s^{2}}[/tex]
Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:
a) t = 50 s.
[tex]\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)[/tex]
[tex]\omega = 50\,\frac{rad}{s}[/tex]
b) t = 100 s.
Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:
[tex]t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}[/tex]
[tex]t = 66.667\,s[/tex]
Since [tex]t > 66.667\,s[/tex], then the angular velocity is equal to zero. Therefore:
[tex]\omega = 0\,\frac{rad}{s}[/tex]
