Answer:
The coefficient of static friction between the object and the disk is 0.087.
Explanation:
According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:
[tex]\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N - m\cdot g = 0[/tex] (2)
Where:
[tex]N[/tex] - Normal force from the ground on the object, measured in newtons.
[tex]m[/tex] - Mass of the object, measured in newtons.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v[/tex] - Linear speed of rotation of the disk, measured in meters per second.
[tex]R[/tex] - Distance of the object from the center of the disk, measured in meters.
By applying (2) on (1), we obtain the following formula:
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s} = \frac{v^{2}}{g\cdot R}[/tex]
If we know that [tex]v = 0.8\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 0.75\,m[/tex], then the coefficient of static friction between the object and the disk is:
[tex]\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}[/tex]
[tex]\mu_{s} = 0.087[/tex]
The coefficient of static friction between the object and the disk is 0.087.
