Question

A blink of an eye is a time interval of about 150ms for an average adult. The closure portion of the blink takes only about 55ms. Let us model the closure of the upper eyelid as uniform angular acceleration through an angular displacement of 16.6 degree. What is the value of the angular acceleration the eyelid undergoes while closing 2. What is the tangential acceleration of the edge of the eyelid while closing if the radius of the eyeball is 1.25 cm?

Answer 1

Explanation:

The first thing is to convert the time into a second, just like this:

`t=55\text{ ms}\cdot\frac{1\text{ s}}{1000\text{ ms}}=0.055\text{ s}`

Now, convert the angular displacement of the eyelid from degrees to rad:

`\partial\theta=16.6\text{\degree}\cdot\frac{2\pi\text{ rad}}{360\text{\degree}}=0.29\text{ rad}`

We can calculate the angular velocity, dividing the angular momentum by the time, like this:

`w=(0.29)/(0.055)=5.27\text{ rad/s}`

The angular acceleration is calculated by means of the quotient of the difference in angular velocity and time, like this:

`a_w=(\delta w)/(\delta t)=(5.27-0)/(0.15-0.055)=55.47\text{ rad/s}^2`

the tangential acceleration would be: