Required information A "blink of an eye" is a time interval of about $150 \mathrm{~ms}$ for an average adult. The "closure" portion of the blink takes only about $55 \mathrm{~ms}$. Let us model the closure of the upper eyelid as uniform angular acceleration through an angular displacement of $126^{\circ}$. What is the tangential acceleration of the edge of the eyelid while closing if the radius of the eyeball is $1.25 \mathrm{~cm}$ ? $\mathrm{m} / \mathrm{s}^{2}$

Step 1 ：Let's denote the angular acceleration as \(\alpha\), the angular displacement as \(\theta\), and the time as \(t\). According to the problem, we have \(\theta = 126^\circ = \frac{7}{4}\pi\) rad, \(t = 55\) ms = \(0.055\) s.

Step 2 ：According to the equation of motion for angular acceleration, we have \(\theta = \frac{1}{2}\alpha t^2\). We can solve this equation for \(\alpha\) to get \(\alpha = \frac{2\theta}{t^2}\).

Step 3 ：Substitute the given values into the equation to get \(\alpha = \frac{2\times\frac{7}{4}\pi}{(0.055)^2} = \frac{7\pi}{0.0015}\) rad/s^2.

Step 4 ：The tangential acceleration \(a_t\) is related to the angular acceleration \(\alpha\) by the equation \(a_t = r\alpha\), where \(r\) is the radius of the circle.

Step 5 ：Substitute the given radius and the calculated angular acceleration into the equation to get \(a_t = 1.25\times10^{-2}\times\frac{7\pi}{0.0015} = \frac{7\pi}{0.012}\) m/s^2.

Step 6 ：Finally, the tangential acceleration of the edge of the eyelid while closing is \(\boxed{\frac{7\pi}{0.012}}\) m/s^2.