Answer:

(a)

\(v\ge\frac13\sqrt{26ag}.\)

(b)

\(R_{\max}=6mg.\)

(a) Let \(w\) be the speed of the particle at the highest point of the circle. For the particle to complete vertical circles on the inner surface, the normal reaction at the highest point must be non-negative.

At the highest point, resolving towards the centre gives

\(R+mg=\frac{mw^2}{a}.\)

Since \(R\ge0\),

\(\frac{mw^2}{a}\ge mg,\)

so

\(w^2\ge ag.\)

From point \(A\) to the highest point, the particle rises a height

\(a-a\cos\theta=a(1-\cos\theta).\)

Using conservation of energy,

\(\frac12mv^2=\frac12mw^2+mga(1-\cos\theta).\)

Thus

\(\frac12mv^2-\frac12mw^2=mga(1-\cos\theta).\)

Since \(w^2\ge ag\),

\(\frac12mv^2\ge \frac12mag+mga(1-\cos\theta).\)

Given \(\cos\theta=\frac1{18}\),

\(1-\cos\theta=1-\frac1{18}=\frac{17}{18}.\)

Therefore

\(\frac12mv^2 \ge \frac12mag+mg a\left(\frac{17}{18}\right).\)

So

\(\frac12mv^2\ge \frac{9}{18}mag+\frac{17}{18}mag =\frac{26}{18}mag =\frac{13}{9}mag.\)

Hence

\(v^2\ge\frac{26}{9}ag,\)

and so

\(v\ge\frac13\sqrt{26ag}.\)

(b) Now

\(v=\frac13\sqrt{26ag},\)

so

\(v^2=\frac{26}{9}ag.\)

The greatest normal reaction occurs at the lowest point, where the speed is greatest. Let the speed at the lowest point be \(V\).

From \(A\) to the lowest point, the particle falls a vertical height

\(a+a\cos\theta=a(1+\cos\theta).\)

By energy conservation,

\(\frac12mv^2+mga(1+\cos\theta)=\frac12mV^2.\)

Substitute \(v^2=\frac{26}{9}ag\) and \(\cos\theta=\frac1{18}\):

\(V^2=\frac{26}{9}ag+2ag\left(1+\frac1{18}\right).\)

Since \(1+\frac1{18}=\frac{19}{18}\),

\(V^2=\frac{26}{9}ag+\frac{19}{9}ag =5ag.\)

At the lowest point, resolving towards the centre gives

\(R_{\max}-mg=\frac{mV^2}{a}.\)

Therefore

\(R_{\max}-mg=\frac{m(5ag)}{a}=5mg.\)

So

\(R_{\max}=6mg.\)