Answer
$\tau = 11.7~N\cdot m$
Work Step by Step
In part (a), we found that the magnitude of the angular acceleration is $~~\alpha = 7.66~rad/s^2$
We can find the rotational inertia:
$I = (\frac{1}{12})(6.40~kg)(1.20~m)^2+(2)(1.06~kg)(0.600~m)^2$
$I = 1.53~kg~m^2$
We can find the magnitude of the torque:
$\tau = I~\alpha$
$\tau = (1.53~kg~m^2)(7.66~rad/s^2)$
$\tau = 11.7~N\cdot m$
