Answer:
0.00321 Nm
Explanation:
[tex]\omega_f[/tex] = Final angular velocity = 3.5 rad/s
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\alpha[/tex] = Angular acceleration
[tex]\theta[/tex] = Angle of rotation = 2 rev
Equation of rotational motion
[tex]\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{3.5^2-0^2}{2\times 2\pi \times 1.8}\\\Rightarrow \alpha=0.54156\ rad/s^2[/tex]
Moment of inertia is given by
[tex]I=mr^2\\\Rightarrow I=0.25\times 0.154^2\\\Rightarrow I=0.005929\ kgm^2[/tex]
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=0.005929\times 0.54156\\\Rightarrow \tau=0.00321\ Nm[/tex]
The torque required is 0.00321 Nm
