Answer
$\tau = 0.0524~N~m$
Work Step by Step
We can find the moment of inertia of the spherical shell plus the two masses on the equator of the shell. Note that the two masses on the axis do not increase the moment of inertia.
$I = \frac{2}{3}MR^2+2mR^2$
$I = \frac{2}{3}(8.40~kg)(0.250~m)^2+(2)(2.00~kg)(0.250~m)^2$
$I = 0.600~kg~m^2$
We can find the change in angular speed.
$\Delta \omega = (50.0~rpm-75.0~rpm)(1~min/60~s)(2\pi~rad/rev)$
$\Delta \omega = -2.618~rad/s$
We can find the required angular deceleration.
$\alpha = \frac{\Delta \omega}{t}$
$\alpha = \frac{-2.618~rad/s}{30.0~s}$
$\alpha = -0.08727~rad/s^2$
We can use the magnitude of the angular deceleration to find the required torque from friction.
$\tau = I \alpha$
$\tau = (0.600~kg~m^2)(0.08727~rad/s^2)$
$\tau = 0.0524~N~m$
