Answer: 0.04
Explanation:
Given
Radius of turntable [tex]r=0.8\ m[/tex]
Block is present at a distance of [tex]r_o=0.4\ m[/tex]
Turntable rotates at a speed of [tex]v=0.8\ m/s[/tex]
angular speed of turntable [tex]\omega =\dfrac{v}{r}[/tex]
[tex]\Rightarrow \omega =\dfrac{0.8}{0.8}\\\Rightarrow \omega =1\ rad/s[/tex]
block will experience a force i.e. centripetal force equal to [tex]m\omega ^2r[/tex]. This must balance the friction force [tex]\mu mg[/tex]
[tex]\Rightarrow m\omega^2 r_o=\mu mg\\\Rightarrow 1^2\times 0.4=\mu \times 9.8\\\Rightarrow \mu =0.04[/tex]
Thus, the coefficient of static friction force is [tex]0.04[/tex]
