Answer:
Explanation:
Given
mass of disk [tex]m=12.5 kg[/tex]
radius of disc [tex]R=0.23 m[/tex]
mass of ring [tex]m_r=7 kg[/tex]
Force [tex]F=9.7 N[/tex]
[tex]N=180 rpm[/tex]
[tex]\omega =\frac{2\pi N}{60}[/tex]
[tex]\omega =6\pi rad/s[/tex]
Total moment of inertia
=Moment of inertia of Disc +Moment of Inertia of ring
[tex]=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2[/tex]
[tex]=13.25\times 0.23^2=0.7009 kg-m^2[/tex]
Now Torque is [tex]T=F\times R=I\cdot \alpha [/tex]
[tex]9.7\times 0.23=0.7\times \alpha [/tex]
[tex]\alpha =3.18 rad/s^2[/tex]
Now using [tex]\omega _f=\omega +\alpha t[/tex]
[tex]\omega _f=0[/tex] here
[tex]0=6\pi -3.18\times t[/tex]
[tex]t=\frac{6\pi }{3.18}[/tex]
[tex]t=5.92 s[/tex]