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:
![\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}](https://tex.z-dn.net/?f=%5Cmu_%7Bs%7D%5Ccdot%20m%5Ccdot%20g%20%3D%20m%5Ccdot%20%5Cfrac%7Bv%5E%7B2%7D%7D%7BR%7D)
![\mu_{s} = \frac{v^{2}}{g\cdot R}](https://tex.z-dn.net/?f=%5Cmu_%7Bs%7D%20%3D%20%5Cfrac%7Bv%5E%7B2%7D%7D%7Bg%5Ccdot%20R%7D)
If we know that
,
and
, then the coefficient of static friction between the object and the disk is:
![\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}](https://tex.z-dn.net/?f=%5Cmu_%7Bs%7D%20%3D%20%5Cfrac%7B%5Cleft%280.8%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%7D%7B%5Cleft%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%280.75%5C%2Cm%29%7D)
![\mu_{s} = 0.087](https://tex.z-dn.net/?f=%5Cmu_%7Bs%7D%20%3D%200.087)
The coefficient of static friction between the object and the disk is 0.087.