Answer:
25 kg m^2/s
Explanation:
The angular acceleration due to the application of torque to the grinding wheel is:
[tex]\tau=I\alpha \rightarrow \alpha=\frac{2.5}{2}=1.25[/tex] (in radian per square second)
So, by the basic equation of the angular acceleration, we can get the angular velocity at t = 10 s as follows:
[tex]\omega=\omega_{o}+\alpha t \rightarrow \omega =0+(1.25)(10) =12.5[/tex] rad/s.
So, the angular momentum at t = 10 s:
[tex]L=I\omega=(2)(12.5) = 25[/tex] kg m^2/s
