Answer:
t = 6.17 s
Explanation:
For a 1 revolution movement, [tex]\triangle \theta = 2\pi[/tex]
Torque, [tex]\tau = 50 Nm[/tex]
Moment of Inertia, [tex]I = 150 kg m^2[/tex]
If the wheel starts from rest, [tex]w_{0} = 0 rad/s[/tex]
The angular displacement of the wheel can be given by the formula:
[tex]\triangle \theta = \omega_0 t + 0.5 \alpha t^2[/tex]................(1)
Where [tex]\alpha[/tex] is the angular acceleration
[tex]\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2[/tex]
To get t, put all necessary parameters into equation (1)
[tex]2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s[/tex]
