Question

A uniform disc of radius $R$ is spinned to the angular velocity $\omega$ and then carefully placed on a horizontal surface. How long will the disc be rotating on the surface if the friction coefficient is equal to $k$ ? The pressure exerted by the disc on the surface can be regarded as uniform.

• A. $t=\frac{5R{\omega }_0}{4kg}$
• B. $t=\frac{3R{\omega }_0}{4kg}$
• C. $t=\frac{3R{\omega }_0}{8kg}$
• D. None of these

Answer 1

The correct option is B $t=\frac{3R{\omega }_0}{4kg}$

It is the moment of friction force which brings the disc to rest. The force of friction is applied to each section of the disc, and since these sections lie at different distance from the axis, the moments of the forces of friction differ from section to section.
To find $N_z$, where $z$ is the axis of rotation of the disc let us partition the disc into thin rings (figure shown below). The force of friction acting on the considered element $dfr=k\left(2\pi rdr\sigma \right)g$, (where $\sigma$ is the density of the disc)
The moment of this force of friction is $d{N}_z=-rdfr=-2\pi k\sigma g{r}^{2}dr$
Integrating w.r.t. $r$ from zero to $R$, we get
$N_z=-2\pi k\sigma g{\int }_0^R{r}^{2}dr=-\frac{2}{3}\pi k\sigma g{R}^3$.
For the rotation of the disc about the stationary axis $z$, from the equation $N_z=I{\beta }_z$
$-\frac{2}{3}\pi k\sigma g{R}^3=\frac{\left(\pi R^2\sigma \right)R^2}{2}{\beta }_z$ or ${\beta }_z=-\frac{4kg}{3R}$
Thus from the angular kinematical equation
${\omega }_z={\omega }_{0z}+{\beta }_{z}t$
$0={\omega }_0+(-\frac{4kg}{3R})t$ or $t=\frac{3R{\omega }_0}{4kg}$