Question 2.14: The coffee cup in Example 2.13 is removed from the drag race...

Part (a) The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal the distance h/2 in Fig. 2.23. Thus
\frac{h}{2}=0.03m=\frac{\Omega^2 R^2}{4g}=\frac{\Omega ^2(0.03 m)^2}{4(9.81 m/s^2)}
Solving, we obtain
\Omega^2=1308 or \Omega=36.2 rad/s = 345 r/min
Part (b)
To compute the pressure, it is convenient to put the origin of coordinates r and z at the bottom of the free-surface depression, as shown in Fig. E2.14. The gage pressure here is P_{0}= 0, and point A is at (r, z) =(3 cm, – 4 cm). Equation (2.46) p=p_{0}-\gamma_{z}+\frac{1}{2}\rho r^2\Omega ^2 can then be evaluated:
p_{A}=0 – (1010 kg/m^3)(9.81 m/s^2)(-0.04 m)+ \frac{1}{2}(1010 kg/m^3)(0.03 m)^2(1308 rad^2/s^2)

=396 N/m^2 + 594 N/m2 = 990 Pa
This is about 43 percent greater than the still-water pressure p_{A}=694 Pa.