A motorcycle rides on the vertical walls around the perimeter of a large circular room. The friction coefficient between the motorcycle tires and the walls is $\mu$. How does the minimum $\mu$ needed to prevent the motorcycle from slipping downwards change with the motorcycle's speed, $s$?
$\textbf{(A)}\mu \propto s^{0} \qquad
\textbf{(B)}\mu \propto s^{-\frac{1}{2}}\qquad    
\textbf{(C)}\mu \propto s^{-1}\qquad
\textbf{(D)}\mu \propto s^{-2}\qquad
\textbf{(E)}\text{none of these}$
1. **Identify the forces acting on the motorcycle:**
   - The gravitational force acting downwards: \( F_g = mg \)
   - The normal force exerted by the wall, which provides the centripetal force: \( N = \frac{mv^2}{r} \)
   - The frictional force acting upwards to counteract gravity: \( F_f = \mu N \)

2. **Set up the equilibrium condition:**
   - For the motorcycle to not slip downwards, the frictional force must balance the gravitational force:
     \[
     F_f = F_g
     \]
     Substituting the expressions for \( F_f \) and \( F_g \):
     \[
     \mu N = mg
     \]

3. **Substitute the normal force \( N \) with the centripetal force expression:**
   \[
   N = \frac{mv^2}{r}
   \]
   Therefore:
   \[
   \mu \frac{mv^2}{r} = mg
   \]

4. **Solve for the coefficient of friction \( \mu \):**
   \[
   \mu = \frac{mg \cdot r}{mv^2}
   \]
   Simplify the expression by canceling out \( m \):
   \[
   \mu = \frac{gr}{v^2}
   \]

5. **Determine the relationship between \( \mu \) and \( v \):**
   - From the equation \( \mu = \frac{gr}{v^2} \), it is clear that:
     \[
     \mu \propto \frac{1}{v^2}
     \]
   - This implies that the minimum coefficient of friction \( \mu \) needed to prevent the motorcycle from slipping downwards is inversely proportional to the square of the speed \( v \).

The final answer is \( \boxed{ \mu \propto s^{-2} } \)