Physics Solution: Rod on Cylinder Equilibrium
1. Physical Analysis
Consider the equilibrium of the rod as it tilts by angle $\theta$ to the right. Due to the condition of rolling without slipping, the contact point $P$ shifts along the rod.
- Geometry of Rolling: The arc length on the cylinder is $s = R\theta$. This length corresponds to the segment of the rod $GP$. Thus, distance $GP = R\theta$.
- Relative Positions:
- The contact point $P$ is at angle $\theta$ to the right of the vertical.
- The center of mass $G$ is “uphill” (to the left) of $P$. This creates a restoring torque that stabilizes the rod.
- The beetle travels to the “downhill” (right) end of the rod. Its weight creates the disturbing torque that tilts the rod.
Figure 1: Corrected Diagram. $G$ is to the left (uphill) providing restoring torque. Beetle is to the right (downhill).
2. Torque Balance Equation
We balance torques about the instantaneous axis of rotation (point $P$).
The Rod’s weight ($mg$) at $G$ tries to rotate the system Counter-Clockwise (back to top).
The Beetle’s weight ($0.2m$) at the end tries to rotate the system Clockwise (further down).
$$ \tau_{rod} = \tau_{beetle} $$ $$ mg \times (\text{lever arm } G) = 0.2m \times (\text{lever arm } B) $$
Lever arms are the horizontal distances from $P$:
- Horizontal dist to $G$: $(PG)\cos\theta = R\theta \cos\theta$
- Horizontal dist to Beetle: $(PB)\cos\theta = (L/2 – R\theta) \cos\theta$
$$ mg (R\theta \cos\theta) = 0.2mg \left( \frac{L}{2} – R\theta \right) \cos\theta $$
Canceling $mg \cos\theta$:
$$ R\theta = 0.2 \left( \frac{L}{2} – R\theta \right) $$ $$ R\theta = 0.1L – 0.2R\theta $$ $$ 1.2 R\theta = 0.1L $$
3. Solving for Parameters
Given $L = 2\pi R$:
$$ 1.2 R\theta = 0.1 (2\pi R) $$ $$ 1.2 \theta = 0.2 \pi \implies \theta = \frac{\pi}{6} = 30^\circ $$
Friction Coefficient
For static equilibrium, the net force along the tangential direction must be zero. The rod tends to slide down to the right, so friction $f$ acts up the slope (left).
Resolving forces along the horizontal:
$$ N \sin\theta = f \cos\theta \implies f = N \tan\theta $$
The condition for no slipping is $f \le \mu N$:
$$ N \tan\theta \le \mu N \implies \mu \ge \tan(30^\circ) $$