## College Physics (4th Edition)

A torque of $~1.48~N\cdot m~$ must be applied.
We can find the rotational inertia of the system about this axis: $I = \sum M_i~R_i^2$ $I = (\sum M_i)~(0.375~m)^2$ $I = (4.0~kg+3.0~kg+5.0~kg+2.0~kg)~(0.375~m)^2$ $I = 1.97~kg~m^2$ We can find the required torque: $\tau = I~\alpha$ $\tau = (1.97~kg~m^2)(0.75~rad/s^2)$ $\tau = 1.48~N\cdot m$ A torque of $~1.48~N\cdot m~$ must be applied.