Answer:
3.3619 Nm
54.27472 rad
182.46618 J
86.88 W
Explanation:
[tex]L_i[/tex] = Initial angular momentum = 7.2 kgm²/s
[tex]L_f[/tex] = Final angular momentum = 0.14 kgm²/s
I = Moment of inertia = 0.142 kgm²
t = Time taken
Average torque is given by
[tex]\tau_{av}=\frac{L_f-L_i}{\Delta t}\\\Rightarrow \tau_{av}=\frac{0.14-7.2}{2.1}\\\Rightarrow \tau_{av}=-3.3619\ Nm[/tex]
Magnitude of the average torque acting on the flywheel is 3.3619 Nm
Angular speed is given by
[tex]\omega_i=\frac{L_i}{I}[/tex]
Angular acceleration is given by
[tex]\alpha=\frac{\tau}{I}[/tex]
From the equation of rotational motion
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=\frac{L_i}{I}\times t+\frac{1}{2}\times \frac{\tau}{I}\times t^2\\\Rightarrow \theta=\frac{7.2}{0.142}\times 2.1+\frac{1}{2}\times \frac{-3.3619}{0.142}\times 2.1^2\\\Rightarrow \theta=54.27472\ rad[/tex]
The angle the flywheel turns is 54.27472 rad
Work done is given by
[tex]W=\tau\theta\\\Rightarrow W=-3.3619\times 54.27472\\\Rightarrow W=-182.46618\ J[/tex]
Work done on the wheel is 182.46618 J
Power is given by
[tex]P=\frac{W}{t}\\\Rightarrow P=\frac{-182.46618}{2.1}\\\Rightarrow P=-86.88\ W[/tex]
The magnitude of the average power done on the flywheel is 86.88 W
