Answer:[tex]\alpha =10.66 rad/s^2[/tex]
Explanation:
Given
mass of disk [tex]m=5 kg[/tex]
diameter of disc [tex]d=30 cm[/tex]
Force applied [tex]F=4 N[/tex]
Now this force will Produce a torque of magnitude
[tex]T=F\cdot r[/tex]
[tex]T=4\dot 0.15[/tex]
[tex]T=0.6 N-m[/tex]
And Torque is given Product of moment of inertia and angular acceleration [tex](\alpha )[/tex]
[tex]T=I\cdot \alpha [/tex]
Moment of inertia for Disc [tex]I= \frac{Mr^2}{2}[/tex]
[tex]I=0.05625 kg-m^2[/tex]
[tex]0.6=0.05625\cdot \alpha [/tex]
[tex]\alpha =10.66 rad/s^2[/tex]
