Question

2. A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating the disk. When the disk reaches a speed of 0.8 m/s, the object starts to slide off the disk. What is the coefficient of static friction between the object and the disk? (4 pts)

Answer 1

Answer:

$\frac{0.065}{r}$

Explanation:

The maximum velocity of an object moving in a curve beyond which it will slide off the curve is given by the relationship in equation (1);

$v=\sqrt{\mu gr}....................(1)$

where $\mu$ is the coefficient of friction between the object and the surface of the curve, g is acceleration due to gravity and r is the radius of the curve.

Given;

v = 0.8m/s

g = $9.81m/s^2$

r = ?

$\mu=?$

In order to solve for $\mu$, we can simply make it the subject of formula from equation (1) as follows;

$v^2=\mu gr\\hence\\\mu=\frac{v^2}{gr}.................(2)$

since we were not given the value of r, we can just substitute other known values, then solve and leave the answer in terms of r.

Therefore;

$\mu=\frac{0.8^2}{9.81r}$

$\mu=\frac{0.64}{9.81r}\\\\\mu=\frac{0.065}{r}$