1. Force Analysis on the Incline

The disc moves in a circle on a plane inclined at angle $\theta$. Key Force Components:

• Gravity ($mg \sin\theta$): Acts constantly “down the slope”. In the diagram, this is the direction towards the bottom edge of the plane.
• Friction ($f_k$): Since the disc is sliding, kinetic friction acts opposite to the instantaneous velocity vector ($\vec{v}$). Thus, friction is always tangential.
• Tension ($T$): Acts radially towards the center (the nail).

Given $\mu = \tan\theta$, the magnitude of kinetic friction is $f_k = \mu N = \mu mg \cos\theta = mg \sin\theta$. Interestingly, $f_k = |F_{gravity\_plane}|$.

2. Solving Part (a): Minimum Tension

Consider the particle at an angle $\phi$ from the lowest point. Tangential Equation: The retarding force is the sum of friction and the tangential component of gravity. $$ F_{ret} = f_k + (mg \sin\theta) \sin\phi $$ Substituting $f_k = mg \sin\theta$: $$ F_{ret} = mg \sin\theta (1 + \sin\phi) $$ Using Work-Energy Theorem: $$ \frac{1}{2}mv^2 – \frac{1}{2}mu^2 = – \int_0^\phi mg \sin\theta (1 + \sin\alpha) l \, d\alpha $$ $$ v^2 = u^2 – 2gl \sin\theta (\phi – \cos\phi + 1) $$ Radial Equation: $$ T – mg \sin\theta \cos\phi = \frac{mv^2}{l} $$ $$ T = mg \sin\theta \cos\phi + \frac{m}{l} [ u^2 – 2gl \sin\theta (\phi – \cos\phi + 1) ] $$ Simplifying the variable parts dependent on $\phi$: $$ T(\phi) \propto \cos\phi – 2(\phi – \cos\phi) = 3\cos\phi – 2\phi $$ To find the minimum, set $\frac{dT}{d\phi} = 0$: $$ -3\sin\phi – 2 = 0 \implies \sin\phi = -\frac{2}{3} $$ Since $\sin\phi$ is negative, the particle has crossed the top point. Let $\alpha$ be the angle past the top. Then $\phi = \pi + \alpha$. $$ \sin(\pi + \alpha) = -\frac{2}{3} \implies -\sin\alpha = -\frac{2}{3} \implies \sin\alpha = \frac{2}{3} $$

3. Solving Part (b): Minimum Velocity

For a complete circle, the particle must traverse the entire path length $2\pi l$. Friction acts constantly along the path. Work done by friction = $-f_k (2\pi l) = – (mg \sin\theta) 2\pi l$. Work done by gravity = 0 (closed loop). $$ \Delta K = W_{total} $$ $$ 0 – \frac{1}{2}mu_{min}^2 = -2\pi l mg \sin\theta $$ $$ u_{min}^2 = 4\pi g l \sin\theta $$ $$ u_{min} = \sqrt{4\pi g l \sin\theta} $$