Answer:
Torque will be 0.2727 N -m
Explanation:
We have given rotational inertia of the motor [tex]I=0.36kgm^2[/tex]
Initial angular speed 25 rad/sec
Final angular speed = 20 rad/sec
Time to change in speed t = 6.6 sec
From first equation of motion [tex]\omega _f=\omega _i+\alpha t[/tex]
[tex]20=25+\alpha \times 6.6[/tex]
[tex]\alpha =-0.7575rad/sec^2[/tex]
Now torque [tex]\tau =I\alpha =0.36\times 0.7575=0.2727N-m[/tex]
