Answer
The torque required is $\tau = 2.1~m\cdot N$. The torque comes from the muscles of David's body acting on the end of the sling in David's hand.
Work Step by Step
$\omega = (75~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = 2.5~\pi~rad/s$
We can find the angular acceleration $\alpha$:
$\alpha = \frac{\omega}{t} = \frac{2.5\pi~rad/s}{5.0~s}$
$\alpha = \frac{\pi}{2.0}~rad/s^2$
We can find the torque required to produce the acceleration:
$\tau = I \alpha = ((0.60~kg)\cdot (1.5~m)^2)( \frac{\pi}{2.0}~rad/s^2)$
$\tau = 2.1~m\cdot N$
The torque required is $\tau = 2.1~m\cdot N$. The torque comes from the muscles of David's body acting on the end of the sling in David's hand.
