Respuesta :
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
[tex]\theta=20^0[/tex]*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
[tex]v_{avg}=\frac{\theta}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} v_{avg}=\frac{20.0^0\times(\frac{\pi}{180})}{71.7\times10^{-3}} \\ =4.86\text{ rad/s} \end{gathered}[/tex]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
[tex]\alpha=\frac{v_{avg}}{t}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \alpha=\frac{4.86}{71.7\times10^{-3}} \\ =67.78rad/s^2 \end{gathered}[/tex]Hence, the angular acceleration of the eye is 67.78 rad/s^2
(C)
The formula for the tangential acceleration is given as
[tex]\alpha_t=\alpha r[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} \alpha_t=(67.78)(0.0125) \\ =0.847m/s^2 \end{gathered}[/tex]Hence, the tangential acceleration is 0.847 m/s^2
