New solution

Tete edited
View solution History
revision #19976 newer →
en/2.1.66.md +25 −0
@@ -0,0 +1,25 @@
+### Statement
+
+$2.1.66.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $2.1.66$ |500x540, 31%](../../img/2.1.66/Savchenko.png)
+
+If the rider uses minimum speed possible, the friction from the surface is maximum possible, i.e. $f=\mu N$. As the vertical component of the total force is zero, we have
+
+\[f\sin\alpha-N\cos\alpha-mg=0\qquad\Rightarrow N(\mu\sin\alpha-\cos\alpha)=mg.\]
+
+The horizontal component of the total force serves as the centripetal force, so
+
+\[f\cos\alpha+N\sin\alpha=\frac{mv^2}{R\sin\alpha}\qquad\Rightarrow N(\mu\cos\alpha+\sin\alpha)=\frac{mv^2}{R\sin\alpha}.\]
+
+Eliminating $N$ from both equations and solving for $v$, we have
+
+\[v=\sqrt{\frac{gR\sin\alpha(\mu\cos\alpha+\sin\alpha)}{\mu\sin\alpha-\cos\alpha}}=\sqrt{\frac{gR\sin\alpha(\mu+\tan\alpha)}{\mu\tan\alpha-1}},\]
+
+which is possible when $\tan\alpha>1/\mu$.
+
+#### Answer
+
+[Insert a concise answer or boxed result]