Example 2 The solution set of the inequality $(m-1) x<\sqrt{4 x-x^{2}}$ about $x$ is $\{x \mid 0<x<2\}$, find the value of the real number $m$.


Solve $4 x-x^{2} \geqslant 0 \Rightarrow 0 \leqslant x \leqslant 4$
For the original inequality, the solution set is $\{x \mid 0<x<2$ ~
Therefore, $m-1 \geqslant 0$ (if $m-1<0$, then the solution set of the inequality would include $\{x \mid 0<x<4\}$ )
Squaring both sides of the inequality gives: $(m-1)^{2} x^{2}<4 x-x^{2}$, which is $x\left(x-\frac{4}{(m-1)^{2}+1}\right)<0$
Thus, $0<x<\frac{4}{(m-1)^{2}+1}$
Combining with the problem, we have $\frac{1}{(m-1)^{2}+1}=2 \Rightarrow(m-1)^{2}=1$
And $m-1 \geqslant 0$. So $m-1=1$
Combining (1)(2), we know that $m=2$ is the solution.