Answer
(a) $15.8Nm$
(b) greater
(c) $2560rpm$
Work Step by Step
(a) We can find the required torque as follows:
$\tau=I\alpha$
$\implies \tau=(\frac{1}{2}mr^2)(\frac{\omega^2}{2\Delta \theta})$
$\implies \tau=\frac{1}{4}(\frac{mr^2\omega^2}{\Delta \theta})$
We plug in the known values to obtain:
$\tau=\frac{1}{4}\frac{(0.755Kg)(0.152m)^2(362\pi/3 rad/s)^2}{12.6\pi rad}$
$\tau=15.8Nm$
(b) We know that the angular speed of the saw blade is directly proportional to the square root of the angular displacement; thus, it will be greater than 1810 rpm.
(c) We can find the required angular speed as follows:
$\omega=\sqrt{\frac{4\tau \Delta \theta}{mr^2}}$
We plug in the known values to obtain:
$\omega=\sqrt{\frac{4(15.8Nm)(6.30\pi rad)}{(0.755Kg)(0.152m)^2}}$
$\omega=2560rpm$
