$1\mathrm{.}10$ In Example $1\mathrm{.}2$ we replaced a cylindrical problem with a linear approximation. The velocity distribution for this flow, taking the cylindrical character into account $($see Prob. $15\mathrm{.}22$ and also $[2],$ p$.91),$ is

$V_{}=(\frac{k^2}{1-k^2})*(\frac{R^2}{r}-r)$

where $R$ is the radius of the outer cylinder, $r$ is the local radius, $k=\frac{r_{\text{i}\text{n}\text{n}\text{e}\text{r}\text{c}\text{y}\text{l}\text{i}\text{n}\text{d}\text{e}\text{r}}}{R},$ and $$ is the angular velocity of the inner cylinder.

$($a$)$ Verify that this distribution shows a zero velocity at the radius of the outer, non$-$moving cylinder and shows $V_{}=kR$ at the surface of the inner, rotating cylinder.

$($b$)$ The shear rate in cylindrical coordinates, for a fluid whose velocity depends only on $r($equivalent to $d\frac{V}{d}y$ in rectangular coordinates$),$ is given by

$=([$ shear rate cylindrical $],[$ coordinates $])=r\frac{d}{dr}(\frac{V_{}}{r})$

Show that for the above velocity distribution, the shear rate at the surface of the inner cylinder is given by

$=(\frac{2}{1-k^2})$

$($c$)$ Show that the shear rate computed by Eq$.1.$AJ using the values in Example $1\mathrm{.}2$ is $\frac{12\mathrm{.}26}{s},$ which is $1\mathrm{.}15$ times the value for the flat approximation in Example $1\mathrm{.}2.$ The manual for the viscometer shown in Fig. $1\mathrm{.}5$ provides formulae equivalent to those in this problem.