Answer:
The constant torque required to stop the disk is 8.6 N-m in clockwise direction .
Explanation:
Let counterclockwise be positive direction and clockwise be negative direction .
Given
Radius of disk , r = 1.33 m
Mass of disc , m = 70.6 kg
Initial angular velocity , [tex]\omega_i =217 rpm[/tex]
Final angular velocity , [tex]\omega_f =0\, rpm[/tex]
Time taken to stop , t = 2.75 min
Let [tex]\alpha[/tex] be the angular acceleration
We know
[tex]\omega _f=\omega _i+\alpha t[/tex]
=>[tex]0=217+2.75\alpha =>\alpha = -78.9\frac{rev}{min^{2}}[/tex]
=>[tex]\alpha =-\frac{78.9\times 2\pi}{60\times 60}\frac{rad}{s^{2}}=-0.138 \frac{rad}{s^{2}}[/tex]
Torque required to stop is given by
[tex]\tau =I\alpha[/tex]
where moment of inertia , [tex]I=\frac{mr^{2}}{2}=\frac{70.6\times 1.33^{2}}{2}kg.m^{2}=62.5 kg.m^{2}[/tex]
=>[tex]\therefore \tau =-0.138\times 62.5\, N.m=-8.6\, N.m[/tex]
Thus the constant torque required to stop the disk is 8.6 N-m in clockwise direction .