Answer:
The torque applied on the plastic disk is 40.97 N-m.
Explanation:
Given that,
Mass of the disc, m = 220 g = 0.22 kg
Diameter of the disk, d = 23 cm
Radius, r = 11.5 cm = 0.115 m
Initial speed of the disk, [tex]\omega_o=0[/tex]
Final speed of the disk, [tex]\omega_f=1800\ rpm=188.49\ rad/s[/tex]
Time, t = 4.6 s
The relation between torque and angular acceleration is given by :
[tex]\tau=I\alpha[/tex]
I is the moment of inertia of the disk
[tex]\tau=I\alpha[/tex]
[tex]\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f}{t}[/tex]
[tex]\tau=\dfrac{0.22\times (0.115)^2}{2}\times \dfrac{188.49}{4.6}[/tex]
[tex]\tau=40.97\ N-m[/tex]
So, the torque applied on the plastic disk is 40.97 N-m. Hence, this is the required solution.
