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Answer:
[tex]21.6\ \text{kg m}^2[/tex]
[tex]3.672\ \text{Nm}[/tex]
[tex]54.66\ \text{revolutions}[/tex]
Explanation:
[tex]\tau[/tex] = Torque = 36.5 Nm
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\omega_f[/tex] = Final angular velocity = 10.3 rad/s
t = Time = 6.1 s
I = Moment of inertia
From the kinematic equations of linear motion we have
[tex]\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2[/tex]
Torque is given by
[tex]\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2[/tex]
The wheel's moment of inertia is [tex]21.6\ \text{kg m}^2[/tex]
t = 60.6 s
[tex]\omega_i[/tex] = 10.3 rad/s
[tex]\omega_f[/tex] = 0
[tex]\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2[/tex]
Frictional torque is given by
[tex]\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}[/tex]
The magnitude of the torque caused by friction is [tex]3.672\ \text{Nm}[/tex]
Speeding up
[tex]\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}[/tex]
Slowing down
[tex]\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}[/tex]
Total number of revolutions
[tex]\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}[/tex]
[tex]\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}[/tex]
The total number of revolutions the wheel goes through is [tex]54.66\ \text{revolutions}[/tex].