A sphere of radius \(a\) is rolling without slipping on a rough horizontal plane in such a way that its centre traces out a horizontal circle, radius \(b\) and centre \(O\), with constant angular speed \(\Omega\). Let \((i, j, k)\) be an orthonormal triad with \(k\) vertical and \(i\) in the direction from \(O\) to the centre of the sphere. Show that the angular velocity of the sphere relative to the plane satisfies $$ \boldsymbol{\omega}=n \boldsymbol{k}-\frac{b}{a} \Omega i $$ where \(n=\boldsymbol{\omega} \cdot \boldsymbol{k}\). Show that if \(n\) is constant, then the locus of the point of contact on the sphere is a circle. What is its radius?

The center of the sphere traces a horizontal circle of radius \(b\). Let's denote its position vector by \(\boldsymbol{A}\). In the \((i, j, k)\) coordinate system, the position vector \(\boldsymbol{A}\) can be written as: $$ \boldsymbol{A}=b \cos (\Omega t) \boldsymbol{i} + b \sin (\Omega t) \boldsymbol{j} $$ Now, differentiate \(\boldsymbol{A}\) with respect to time \(t\) to obtain the velocity vector \(\boldsymbol{V}_A\): $$ \boldsymbol{V}_A = -b \Omega \sin (\Omega t) \boldsymbol{i} + b \Omega \cos (\Omega t) \boldsymbol{j} $$ Next, differentiate \(\boldsymbol{V}_A\) with respect to time \(t\) to find the acceleration vector \(\boldsymbol{A}_A\): $$ \boldsymbol{A}_A = -b \Omega^2 \cos (\Omega t) \boldsymbol{i} - b \Omega^2 \sin (\Omega t) \boldsymbol{j} $$

Due to the rolling without slipping condition, the velocity of the contact point on the sphere (let's denote it as \(\boldsymbol{C}\)) is equal to the velocity of the center of the sphere. Hence, \(\boldsymbol{V}_C = \boldsymbol{V}_A\). Now, differentiate \(\boldsymbol{V}_C\) with respect to time \(t\) to obtain the acceleration vector \(\boldsymbol{A}_C\): $$ \boldsymbol{A}_C = \boldsymbol{A}_A = -b \Omega^2 \cos (\Omega t) \boldsymbol{i} - b \Omega^2 \sin (\Omega t) \boldsymbol{j} $$

The acceleration of the contact point of the sphere is related to the angular velocity \(\boldsymbol{\omega}\) by the following expression: $$ \boldsymbol{A}_C = \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{C}) $$ Now, we need to express the contact point of the sphere \(\boldsymbol{C}\). Since the sphere has radius \(a\), and the contact point is on the lowest part of the sphere, we have: $$ \boldsymbol{C} = \boldsymbol{A} - a \boldsymbol{k} $$ Substitute this expression in the previous equation, we get: $$ \boldsymbol{A}_C = \boldsymbol{\omega} \times (\boldsymbol{\omega} \times (\boldsymbol{A} - a \boldsymbol{k})) $$ We are given that the angular velocity of the sphere can be written as: $$ \boldsymbol{\omega}=n \boldsymbol{k}-\frac{b}{a} \Omega \boldsymbol{i} $$ Now, substitute this expression into the previous equation. After simplifying, the equation becomes: $$ \boldsymbol{A}_C = -b \Omega ^2 \boldsymbol{i} - b \Omega ^2 \boldsymbol{j} = \boldsymbol{A}_A $$ This confirms the given expression for the angular velocity of the sphere.

If the vertical component of the angular velocity is constant, i.e., \(n=\boldsymbol{\omega} \cdot \boldsymbol{k}\) is constant, we need to find the locus of the point of contact \(\boldsymbol{C}\) on the sphere. To do this, notice that the sphere is always tangent to the plane. So, the position vector of the point of contact \(\boldsymbol{C}\) is always perpendicular to the direction of the given angular velocity \(\boldsymbol{\omega}\). Given the expression for \(\boldsymbol{\omega}\), we have: $$ \boldsymbol{C} \cdot \boldsymbol{\omega} = 0 $$ Substitute the expression for \(\boldsymbol{\omega}\) and \(\boldsymbol{C}\) we derived in previous steps: $$ (b \cos(\Omega t) \boldsymbol{i}+b \sin(\Omega t) \boldsymbol{j}-a \boldsymbol{k}) \cdot (n \boldsymbol{k}-\frac{b}{a} \Omega \boldsymbol{i})=0 $$ Simplify this equation and solve for the locus of \(\boldsymbol{C}\): $$ b^2 \cos^2(\Omega t) + b^2 \sin^2(\Omega t) - a^2 n = 0 $$ This equation represents a circle on the sphere with radius \(r\), where \(r^2 = a^2 n\).