Answer:
109656.25 Nm
Explanation:
[tex]\omega_f[/tex] = Final angular velocity = 1.5 rad/s
[tex]\omega_i[/tex] = Initial angular velocity = 0
[tex]\alpha[/tex] = Angular acceleration
t = Time taken = 6 s
m = Mass of disk = 29000 kg
r = Radius = 5.5 m
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2[/tex]
Torque is given by
[tex]\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm[/tex]
The torque specifications must be 109656.25 Nm