Solution 10: Friction on Disc
1. Forces:
Ball Force $F = \frac{GmM}{R^2}$. Disc pull $mg$. Sliding force $f_s = F \cos \theta$. Normal force $N = mg – F \sin \theta$.
Ball Force $F = \frac{GmM}{R^2}$. Disc pull $mg$. Sliding force $f_s = F \cos \theta$. Normal force $N = mg – F \sin \theta$.
2. Friction Condition:
\[ \mu \ge \frac{f_s}{N} = \frac{F \cos \theta}{mg – F \sin \theta} = \frac{\cos \theta}{A – \sin \theta} \quad \text{where } A = \frac{gR^2}{GM} \]
\[ \mu \ge \frac{f_s}{N} = \frac{F \cos \theta}{mg – F \sin \theta} = \frac{\cos \theta}{A – \sin \theta} \quad \text{where } A = \frac{gR^2}{GM} \]
3. Optimization:
Max value of RHS is $\frac{1}{\sqrt{A^2-1}}$. \[ \mu \ge \frac{1}{\sqrt{\left(\frac{gR^2}{GM}\right)^2 – 1}} = \frac{GM}{\sqrt{(gR^2)^2 – (GM)^2}} \]
Max value of RHS is $\frac{1}{\sqrt{A^2-1}}$. \[ \mu \ge \frac{1}{\sqrt{\left(\frac{gR^2}{GM}\right)^2 – 1}} = \frac{GM}{\sqrt{(gR^2)^2 – (GM)^2}} \]
Answer: (c)