A horizontal cylinder closed from one end is rotated with a constant angular velocity `omega` about a vertical axis passing through the open end of the cylinder. The outside air pressure is equal to `p_0`, the temperature to `T`, and the molar mass of air to `M`. find the air pressure as a function of the distance `r` from the rotate axis. The molar mass is assumed to be independent of `r`.

Explanation

Let us consider the mass element of the gas (thin layer) in the cylinder at a distance `r` from its open end as shown in the figure.
Using Newton's second law for the element
`F_n = mw_n` :
`(p + dp) S - pS = (rho S dr) omega^2 r`
or, `dp = rho omega^2 r dr = (pM)/(RT) omega^2 r dr` So, `(dp)/(p) = (M omega^2)/(RT) r dr` or, `int_(p_0)^p (dp)/(p) = (M omega^2)/(RT) int_0^r r dr`,
Thus, `1n (p)/(p_0) = (M omega^2)/(2RT)r^2` or, `p = p_0 e^(M omega^2 r^2//RT)`.
.