Question

The ability of the human eye to rapidly rotate was studied using contact lenses fitted with accelerometers. While a subject, whose eyeball has radius 1.25 cm, watches a moving object her eyeball rotates through 20.0° in a time interval of 71.7 ms.A. What is the magnitude of the average angular velocity of the eye? B, Assume that the eye starts at rest, rotates with a constant angular acceleration during the first half of the interval, and then the rotation slows with a constant angular acceleration during the second half until it comes to rest. What is the magnitude of the angular acceleration of the eye?C. What tangential acceleration would the contact-lens accelerometers record in this case?

Answer 1

Given data

*The given radius of the eyeball is r = 1.25 cm = 0.0125 m

*The initial angular velocity of the eye is 0 rad/s

*The eyeball rotates at an angle is

`\theta=20^0`

*The given time interval is t = 71.7 ms = 71.7 × 10^-3 s

(A)

The formula for the magnitude of the average angular velocity is given as

`v_(avg)=(\theta)/(t)`

Substitute the known values in the above expression as

`\begin{gathered} v_(avg)=(20.0^0*((\pi)/(180)))/(71.7*10^(-3)) \\ =4.86\text{ rad/s} \end{gathered}`

Hence, the magnitude of the average angular velocity is v_avg = 4.86 rad/s

(B)

The formula for the magnitude of the angular acceleration of the eye is given by the equation of motion as

`\alpha=(v_(avg))/(t)`

Substitute the known values in the above expression as

`\begin{gathered} \alpha=(4.86)/(71.7*10^(-3)) \\ =67.78rad/s^2 \end{gathered}`

Hence, the angular acceleration of the eye is 67.78 rad/s^2

(C)

The formula for the tangential acceleration is given as

`\alpha_t=\alpha r`

Substitute the known values in the above expression as

`\begin{gathered} \alpha_t=(67.78)(0.0125) \\ =0.847m/s^2 \end{gathered}`

Hence, the tangential acceleration is 0.847 m/s^2