Answer:
a) [tex]I = 19.799\,kg\cdot m^{2}[/tex], b) [tex]T = -3.405\,N\cdot m[/tex], c) [tex]n_{T} \approx 54.842\,rev[/tex]
Explanation:
a) The net torque is:
[tex]T = I\cdot \alpha[/tex]
Let assume a constant angular acceleration, which is:
[tex]\alpha = \frac{\omega-\omega_{o}}{t}[/tex]
[tex]\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}[/tex]
[tex]\alpha = 1.793\,\frac{rad}{s^{2}}[/tex]
The moment of inertia of the wheel is:
[tex]I = \frac{T}{\alpha}[/tex]
[tex]I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }[/tex]
[tex]I = 19.799\,kg\cdot m^{2}[/tex]
b) The deceleration of the wheel is due to the friction force. The deceleration is:
[tex]\alpha = \frac{\omega-\omega_{o}}{t}[/tex]
[tex]\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}[/tex]
[tex]\alpha = - 0.172\,\frac{rad}{s^{2}}[/tex]
The magnitude of the torque due to friction:
[tex]T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )[/tex]
[tex]T = -3.405\,N\cdot m[/tex]
c) The total angular displacement is:
[tex]\theta_{T} = \theta_{1} + \theta_{2}[/tex]
[tex]\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}[/tex]
[tex]\theta_{T} = 344.580\,rad[/tex]
The total number of revolutions of the wheel is:
[tex]n_{T} = \frac{\theta_{T}}{2\pi}[/tex]
[tex]n_{T} = \frac{344.580\,rad}{2\pi}[/tex]
[tex]n_{T} \approx 54.842\,rev[/tex]
