Answer:
0.186 N-m
Explanation:
mass of the grindstone, [tex]m=1.7 kg[/tex]
radius, [tex]r=8 cm[/tex]
Frequency, [tex]f=730 rev/min = 12.16 rev/s[/tex]
time, [tex]t=9s[/tex]
final angular velocity, [tex]\omega=0[/tex]
Initial angular velocity,
[tex]\omega_o=2\pi f\\=2\pi (12.16) rad/s\\= 76.36 rad/s[/tex]
Angular acceleration of the grind stone is:
[tex]\alpha=\frac{\omega-\omega_o}{t}\\\Rightarrow \alpha =\frac{0-76.36}{9} = -8.48 rad/s^2[/tex]
Moment of inertia:
[tex]I=mr^2+mr^2=2mr^2[/tex]
[tex]I=2\times 1.7 kg\times (0.08m)^2= 0.022kg-m^2[/tex]
Torque exerted by the ax on the grind stone is:
[tex]\tau=I\alpha\\\tau=0.022\times (-8.48) \\\tau=0.186N-m[/tex]
