# Chapter 5 - Problems - Page 188: 52

(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 $3.1~m/s^2$

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