Answer:
[tex]a=0.276[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=98.kg[/tex]
Radius [tex]r=0.335[/tex]
Angular velocity [tex]\omega=100rpm[/tex]
Radial force of [tex]F_r=23.6 N.[/tex]
Kinetic coefficient of friction [tex]\mu=0.192[/tex]
Generally the equation for Kinetic Force is mathematically given by
[tex]F_k=\mu.F_r[/tex]
[tex]F_k=0.192*23.6[/tex]
[tex]F_k=4.5312[/tex]
Generally the equation for Torque on Center is mathematically given by
[tex]Ia=f_k*r[/tex]
Where
[tex]I=\frac{Mr^2}{2}[/tex]
Therefore
[tex]a=\frac{2f_k}{Mr}[/tex]
[tex]a=\frac{2*4.5312}{98*0.335}[/tex]
[tex]a=0.276[/tex]
Therefore Angular acceleration of the grindstone is
[tex]a=0.276[/tex]