Answer:
Explanation:
Given mass of grindstone [tex]m=90\ kg[/tex]
radius of stone [tex]r=0.34\ m[/tex]
angular speed of disc [tex]\omega =90\ rpm[/tex]
Steel Axle applying a force of [tex]F=20\ N[/tex]
coefficient of kinetic friction [tex]\mu =0.2[/tex]
Frictional Torque applied by steel is given by
[tex]\tau=r\times f_r[/tex]
[tex]\tau =r\times \mu F[/tex]
where [tex]f_r[/tex]=frictional force
[tex]\tau =r\times \mu \times F[/tex]
[tex]\tau =0.34\times 0.2\times 20[/tex]
[tex]\tau =1.36\ N-m[/tex]
Torque is also given by
[tex]\tau =I\cdot \alpha [/tex]
where [tex]\alpha [/tex]=angular acceleration
I=moment of Inertia
[tex]\tau =0.5Mr^2\times \alpha [/tex]
[tex]0.5Mr^2\times \alpha =1.36\ N-m[/tex]
[tex]0.5\times 90\times 0.34^2\times \alpha =1.36[/tex]
[tex]\alpha =0.261\ rad/s^2[/tex]
