Answer
a) $\tau=-0.25N.m$
b) decreases
Work Step by Step
(a) We can find the required torque as follows:
$I=\frac{1}{2}MR^2=\frac{1}{2}(6.4)(0.71)^2=1.6Kgm^2$
$\alpha=\frac{\omega^2-\omega_{\circ}^2}{2\Delta \theta}$
$\alpha=\frac{(0)^2-(1.22)^2}{2(0.75rev\times 2\pi rad/rev)}=-0.158rad/s^2$
Now $\tau=I\alpha$
We plug in the known values to obtain:
$\tau=(1.6)(-0.158)$
$\tau=-0.25N.m$
(b) We know that if the mass of the wheel is doubled and its radius is halved then the moment of inertial will be reduced to half and as result the angle through which the wheel rotates will decrease.