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:
(1)
(2)
Where:
- Normal force from the ground on the object, measured in newtons.
- Mass of the object, measured in newtons.
- Gravitational acceleration, measured in meters per square second.
- Linear speed of rotation of the disk, measured in meters per second.
- Distance of the object from the center of the disk, measured in meters.
By applying (2) on (1), we obtain the following formula:


If we know that
,
and
, then the coefficient of static friction between the object and the disk is:


The coefficient of static friction between the object and the disk is 0.087.