Answer
(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.
Work Step by Step
(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.
