Solution to Question 24
Consider a sand particle on the disc at a distance $r$ from the center. The disc is inclined at an angle $\theta$. The forces acting on the particle along the plane of the disc are:
- The component of gravity pulling it down the slope: $mg \sin \theta$.
- The centrifugal force pushing it radially outward: $m \omega^2 r$.
- The static friction force $f_s$ which opposes the tendency to slip.
Slipping will first occur at the lowest point of the circular path on the disc. At this point, the component of gravity acts radially outward (away from the center), aligning with the centrifugal force. The tendency to slip is maximum here.
The condition for the particle to remain on the disc is that the total outward force must be balanced by the maximum static friction: $$ m \omega^2 r + mg \sin \theta \le f_{\text{max}} $$ $$ m \omega^2 r + mg \sin \theta \le \mu N $$ Since $N = mg \cos \theta$, the limiting condition for slipping is: $$ m \omega^2 r + mg \sin \theta = \mu mg \cos \theta $$ $$ \omega^2 r = g (\mu \cos \theta – \sin \theta) \quad \dots(1) $$
Sand falls off starting from the outer edge where the centrifugal force is highest. If $\eta$ fraction of the sand has fallen off, the sand remains on the inner portion of the disc up to a radius $r$. The area of the original sand layer is $A = \pi R^2$. The area of the remaining sand is $A’ = \pi r^2$. The fraction of sand lost is: $$ \eta = \frac{\pi R^2 – \pi r^2}{\pi R^2} = 1 – \frac{r^2}{R^2} $$ $$ \frac{r^2}{R^2} = 1 – \eta \implies r = R \sqrt{1 – \eta} $$
Substitute this value of $r$ into equation (1) to find the angular velocity $\omega$: $$ \omega^2 (R \sqrt{1 – \eta}) = g (\mu \cos \theta – \sin \theta) $$ $$ \omega = \sqrt{\frac{g (\mu \cos \theta – \sin \theta)}{R \sqrt{1 – \eta}}} $$