Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.5 - Equations of Motion: Normal and Tangential Coordinates - Problems - Page 147: 58

$v=0.969m/s$

We can determine the required speed as $\rho=0.25(\frac{4}{5})=0.2m$ We know that $\Sigma F_n=ma_n$ $\implies N_s(\frac{3}{5})-0.2N_s(\frac{4}{5})=2(\frac{v^2}{\rho})$ $\implies 0.0444N_s=v^2~~~$eq(1) We also know that $\Sigma F_b=ma_b$ $\implies N_s(\frac{4}{5})+0.2N_s(\frac{3}{5})=mg$ $\implies 0.92N_s=2\times 9.81$ $\implies N_s=21.32N$ We plug in this value into eq(1) to obtain: $0.044\times 21.32=v^2$ $\implies v=0.969m/s$