Answer
(a) The average angular velocity is $4.65~rad/s$
(b) The magnitude of the angular acceleration is $248~rad/s^2$
(c) The tangential acceleration is $3.1~m/s^2$
Work Step by Step
(a) We can express the angular displacement in units of radians:
$\Delta \theta = 20.0^{\circ}\times \frac{\pi~rad}{180^{\circ}} = 0.349~rad$
We can find the average angular velocity:
$\omega_{ave} = \frac{\Delta \theta}{t}$
$\omega_{ave} = \frac{0.349~rad}{75\times 10^{-3}~s}$
$\omega_{ave} = 4.65~rad/s$
The average angular velocity is $4.65~rad/s$
(b) During the first half of the time interval, the eyeball rotates through $10.0^{\circ}$ in half the total time. We can find the angular acceleration:
$\Delta \theta = \frac{1}{2}\alpha~t^2$
$\alpha = \frac{2\Delta \theta}{t^2}$
$\alpha = \frac{(2)(0.1745~rad)}{(37.5\times 10^{-3}~s)^2}$
$\alpha = 248~rad/s^2$
The magnitude of the angular acceleration is $248~rad/s^2$
(c) We can find the tangential acceleration:
$a = \alpha~r$
$a = (248~rad/s^2)(0.0125~m)$
$a = 3.1~m/s^2$
The tangential acceleration is $3.1~m/s^2$
