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Answer :
Explanation:
Given that,
Torque = 35.9 N-m
Time of acceleration= 6.20 s
Initial angular speed = 0 rad/s
Final angular speed = 9.9 rad/s
Time of declaration = 60.7 s
(a). We need to calculate the angular acceleration
Using formula of angular acceleration
[tex]\omega_{2}=\omega_{1}+\alpha t_{1}[/tex]
Put the value into the formula
[tex]9.9=0+\alpha\times6.2[/tex]
[tex]\alpha=\dfrac{9.9}{6.2}[/tex]
[tex]\alpha=1.59\ rad/s^2[/tex]
We need to calculate the moment of inertia of the wheel
Using formula of moment of inertia
[tex]I=\dfrac{\tau}{\alpha}[/tex]
Put the value into the formula
[tex]I=\dfrac{35.9}{1.59}[/tex]
[tex]I=22.57\ kg-m^2[/tex]
The moment of inertia of the wheel is [tex]22.57\ kg-m^2[/tex]
(b). The applied force is removed, then the magnitude of the torque due to friction
We need to the magnitude of the torque due to friction
Using equation of rotation
[tex]\omega_{2}=\omega_{1}+\alpha' t_{2}[/tex]
Put the value into the formula
[tex]0=9.9+\alpha'\times60.7[/tex]
[tex]\alpha'=-\dfrac{9.9}{60.7}[/tex]
[tex]\alpha'=-0.163\ rad/s^{2}[/tex]
Magnitude of angular acceleration is 0.163 rad/s²
We need to calculate the frictional torque
Using formula of torque
[tex]\tau'=I\times\alpha'[/tex]
Put the value into the formula
[tex]\tau'=22.57\times 0.163[/tex]
[tex]\tau'=3.678\ kg-m rad/s^2[/tex]
The frictional torque is [tex]3.678\ kg-m\ rad/s^2[/tex]
(c). We need to calculate the total number of revolutions of the wheel during the entire interval of 66.9 s
Using formula of angular displacement
During the applied force,
[tex]\theta_{1}=\dfrac{1}{2}\alpha t_{1}^2[/tex]
Put the value into the formula
[tex]\theta_{1}=\dfrac{1}{2}\times1.58\times6.2^2[/tex]
[tex]\theta_{1}=30.36\ rad[/tex]
During the removed of force,
[tex]\theta_{2}=\omega t^2+\dfrac{1}{2}\alpha' t_{2}^2[/tex]
Put the value into the formula
[tex]\theta_{2}=9.9\times60.7-\dfrac{1}{2}\times0.163\times(60.7)^2[/tex]
[tex]\theta_{2}=300.6\ rad[/tex]
The total angular displacement
[tex]\theta=\theta_{1}+\theta_{2}[/tex]
[tex]\theta=30.36+300.6[/tex]
[tex]\theta=330.96\ rad[/tex]
[tex]\theta=\dfrac{330.96}{2\pi}[/tex]
[tex]\theta=52.67\ rev[/tex]
The total number of revolutions of the wheel is 52.67 rev.
Hence, This is the required solution.
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