Answer:
[tex]\mu = 0.28 [/tex]
Explanation:
Time period of the revolution of the disc is given as
[tex]T = 1.60 s[/tex]
now the angular speed of the disc is given as
[tex]\omega = \frac{2\pi}{T}[/tex]
so we will have
[tex]\omega = \frac{2\pi}{1.60}[/tex]
[tex]\omega = 3.92 rad/s[/tex]
now at the rim position of the disc the net force on the penny is due to friction force
So we will have
[tex]\mu mg = m\omega^2 r[/tex]
here we will have
[tex]\mu g = \omega^2 r[/tex]
[tex]\mu = \frac{\omega^2 r}{g}[/tex]
[tex]\mu = \frac{3.92^2 (0.179)}{9.81}[/tex]
[tex]\mu = 0.28 [/tex]
