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A blink or an eye" is a time interval of about $150 \mathrm{~ms}$ ror an average odut. The talosure porion of me blink takes ony about $55 \mathrm{~ms}$. Let us model the closure of the upper eyelid as uniform angular acceleraton uhrough an angulith. displacement of $17.0^{*}$,
What is the value of the angular acceleration the eyelld uncergoes while closing?
$\mathrm{rad} / \mathrm{s}^{2}$

Step 1 ：The problem is asking for the angular acceleration of the eyelid while closing. We know that the eyelid closes in 55 ms and undergoes an angular displacement of 17.0 degrees. We can use the equation of motion for angular acceleration to solve this problem.

Step 2 ：The equation of motion for angular acceleration is given by: \(\theta = \omega_{i}t + \frac{1}{2}\alpha t^{2}\) where \(\theta\) is the angular displacement, \(\omega_{i}\) is the initial angular velocity, \(t\) is the time, and \(\alpha\) is the angular acceleration.

Step 3 ：In this case, the initial angular velocity is 0 (since the eyelid starts from rest), so the equation simplifies to: \(\theta = \frac{1}{2}\alpha t^{2}\)

Step 4 ：We can rearrange this equation to solve for \(\alpha\): \(\alpha = \frac{2\theta}{t^{2}}\)

Step 5 ：We need to convert the angular displacement from degrees to radians and the time from milliseconds to seconds before substituting into the equation. The conversion gives us \(\theta\) in radians as approximately 0.29670597283903605 and \(t\) in seconds as 0.055.

Step 6 ：Substituting these values into the equation gives us \(\alpha\) as approximately 196.169238240685.

Step 7 ：Final Answer: The angular acceleration the eyelid undergoes while closing is approximately \(\boxed{196.17 \, \mathrm{rad/s^2}}\).