Solution to Question 32
We analyze the block of mass $m$ in the frame of the rotating cylinder. We define the position by the azimuthal angle $\phi$, where $\phi=0$ is the lowest point and $\phi=\pi$ is the highest point of the cross-section.
Forces acting on the block:
- Normal Force ($N$): Acts radially inward to provide centripetal acceleration and balance the normal component of gravity. $$ N = m\omega^2 r + mg \cos\theta \cos\phi $$
- Shear (Tangential) Force ($F_t$): The net force along the surface trying to cause slipping. It has two perpendicular components:
- Axial component (down the incline): $mg \sin\theta$
- Tangential component (along the circle): $mg \cos\theta \sin\phi$
To prevent slipping, friction must be sufficient: $\mu N \ge F_t$.
$$ \mu (m\omega^2 r + mg \cos\theta \cos\phi) \ge mg \sqrt{\sin^2\theta + \cos^2\theta \sin^2\phi} $$We rearrange this to isolate the required angular velocity term:
$$ \frac{\mu \omega^2 r}{g} \ge \underbrace{\sqrt{\sin^2\theta + \cos^2\theta \sin^2\phi} – \mu \cos\theta \cos\phi}_{y(\phi)} $$We need to find the maximum value of $y(\phi)$ to ensure the condition holds for all $\phi$.
If $\mu$ is small, the most critical point is at the very top of the cylinder ($\phi = \pi$). Here, gravity opposes the normal force (reducing friction) and the tangential force is purely axial.
- At $\phi = \pi$, $\cos\phi = -1$ and $\sin\phi = 0$.
- $y(\pi) = \sqrt{\sin^2\theta} – \mu \cos\theta (-1) = \sin\theta + \mu \cos\theta$.
The condition becomes:
$$ \frac{\mu \omega^2 r}{g} \ge \sin\theta + \mu \cos\theta $$ $$ \omega \ge \sqrt{\frac{g(\sin\theta + \mu \cos\theta)}{\mu r}} $$If $\mu$ is large, the block is secure at the top. The critical point shifts to a location where the shear force is larger. We find the maximum of $y(\phi)$ by setting $y'(\phi) = 0$.
$$ \frac{dy}{d\phi} = \frac{\cos^2\theta \sin\phi \cos\phi}{\sqrt{\sin^2\theta + \cos^2\theta \sin^2\phi}} + \mu \cos\theta \sin\phi = 0 $$Solving for the critical angle $\phi_c$ (where $\sin\phi \neq 0$):
$$ \frac{\cos\theta \cos\phi_c}{\sqrt{\sin^2\theta + \cos^2\theta \sin^2\phi_c}} = -\mu $$Squaring and solving for $\cos\phi_c$ yields $\cos^2\phi_c = \frac{\mu^2}{\cos^2\theta(1+\mu^2)}$. Using the negative root (from the derivative equation):
$$ \cos\phi_c = \frac{-\mu}{\cos\theta \sqrt{1+\mu^2}} $$Substituting this back into $y(\phi)$ gives the maximum value:
$$ y_{max} = \sqrt{1+\mu^2} $$Thus the condition becomes:
$$ \frac{\mu \omega^2 r}{g} \ge \sqrt{1+\mu^2} \implies \omega \ge \sqrt{\frac{g\sqrt{1+\mu^2}}{\mu r}} $$