91Ó°ÊÓ

The angular acceleration of any moving body can be described as the changing rate of angular velocity of the body with respect to the time. The unit of angular acceleration is radians per square second.

The angular acceleration of the gritstone will be by the formula of motion,

${\mathrm{Ó\neg }}_z={\mathrm{Ó\neg }}_{0}z+{\mathrm{Î\pm }}_{z}t$ ….. (1)

localid="1664081898094" $$\begin{gathered} ∑{\mathrm{Ï„}}_z=I{\mathrm{Î\pm }}_z \\ {f}_k={\mathrm{μ}}_{k}n \\ {\mathrm{μ}}_k=-\left(MR{\mathrm{Î\pm }}_z\right)/2n…..\left(2\right) \end{gathered}$$

Where, $n$is force, ${\mathrm{Ó\neg }}_z$final rotation speed, ${\mathrm{Ó\neg }}_{0z}$starting rotation speed, ${\mathrm{Î\pm }}_z$rotation acceleration, ${\mathrm{Ï„}}_z$is the torque, ${\mathrm{μ}}_k$coefficient of friction, $f_k$friction force, $I$moment of inertia, and $t$ is time. M is the mass and R is the radius of the object.

Consider the given data as below.

Diameter of disk, $d=0.520m$

Mass of disk, $M=50.0kg$

Rotation, ${\mathrm{Ó\neg }}_{0z}=850rev/min$

Force, $n=160N$

Time, $t=7.50s$

$$\begin{gathered} {\mathrm{Ó\neg }}_z={\mathrm{Ó\neg }}_{0}z+{\mathrm{Î\pm }}_{z}t \\ With{\mathrm{Ó\neg }}_z=0 \end{gathered}$$

$$\begin{gathered} 0=850rev/min+{\mathrm{Î\pm }}_{z}7.50s \\ {\mathrm{Î\pm }}_z=\left(\right(850rev/\left(60sec\right)\left(\right)(2\mathrm{Ï€}rad/1rev)\left)\right)/7.50s \\ {\mathrm{Î\pm }}_z=11.9rad/s^2 \end{gathered}$$

And further, substitute known values into equation (2).

$$\begin{gathered} {\mathrm{μ}}_k=\frac{MR{\mathrm{Î\pm }}_z}{2n} \\ =\frac{-(50.0kg)(0.260m)(-11.9rad/s^2)}{2\left(160N\right)} \\ =0.483 \end{gathered}$$

The coefficient of the friction will be ${\mathrm{μ}}_k=0.483$.