## College Physics (4th Edition)

(a) The angular speed just before release is $13.5~rad/s$ (b) The athlete must apply a torque of $15.6~N \cdot m$ (c) The discus lands a distance of 15.1 meters away.
(a) We can find the angular speed just before release: $\Delta \theta = \omega_{ave}~t$ $\Delta \theta = \frac{\omega_0+\omega_f}{2}~t$ $\omega_f = \frac{2~\Delta \theta}{t}-\omega_0$ $\omega_f = \frac{(2)(1.5)(2\pi~rad)}{1.4~s}-0$ $\omega_f = 13.5~rad/s$ The angular speed just before release is $13.5~rad/s$ (b) We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{13.5~rad/s-0}{1.4~s}$ $\alpha = 9.64~rad/s^2$ We can find the torque: $\tau = I~\alpha$ $\tau = MR^2~\alpha$ $\tau = (2.0~kg)(0.90~m)^2~(9.64~rad/s^2)$ $\tau = 15.6~N \cdot m$ The athlete must apply a torque of $15.6~N \cdot m$ (c) We can find the speed of he discus when it is released: $v = \omega~r = (13.5~rad/s)(0.90~m) = 12.15~m/s$ We can assume that the the vertical height of the discus when it is released and when it lands is approximately equal. We can find the horizontal distance the discus travels: $x = \frac{v^2~sin(2\theta)}{g}$ $x = \frac{(12.15~m/s)^2~sin((2)(45^{\circ}))}{9.80~m/s^2}$ $x = 15.1~m$ The discus lands a distance of 15.1 meters away.