Answer:
The torque must be applied to the wheel is 15.7 N-m.
Explanation:
Given that,
Mass of the wheel of cylinder, m = 50 kg
Diameter of the wheel, d = 1 m
Radius, r = 0.5 m
Initial speed off the wheel is 0
Final angular speed of the wheel, 120 rev/min = 12.56 rad/s
Angular displacement, [tex]\theta=20\ rev=125.6\ rad[/tex]
The torque is given by :
[tex]\tau=I\alpha \\\\\tau=\dfrac{mr^2}{2}\times (\dfrac{\omega_f^2}{2\theta})\\\\\tau=\dfrac{50\times1^{2}}{2}\times(\dfrac{12.56^{2}}{2\times125.6})\\\\\tau=15.7\ N-m[/tex]
So, the torque must be applied to the wheel is 15.7 N-m.
