Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 148: 67

$\theta=26.7^{\circ}$

We can determine the required tilt angle as follows: $\Sigma F_b=0$ $\implies Ncos\theta-mg=0$ $\implies N=\frac{mg}{cos\theta}$ We know that $\Sigma F_n=ma_n$ $\implies Nsin\theta=m\frac{v^2}{\rho}$ $\implies \frac{mgsin\theta}{cos\theta}=m\frac{v^2}{\rho}$ We plug in the known values to obtain: $9.81\times tan\theta=\frac{22.22m/s}{100}$ This simplifies to: $\theta=26.7^{\circ}$