The net torque, #tau#, can be expressed in terms of the moment of inertia and angular acceleration, where #tau=Ialpha#.
Assuming that this is a thin, rigid rod, with the axis of rotation through its center, the moment of inertia, #I#, of the rod is given by #1/12ML^2#, where #M# is the mass of the rod and #L# is its length. If the rod is rotated about its end, #I=1/3ML^2#. I will show the calculation for the axis of rotation through the center. We are given both #L# and #M#. Thus,
#I=1/12*2kg*(5m)^2#
#I=50/12kgm^2#
The angular acceleration of the rod as it rotates, #alpha#, can be calculated from the given values of time and frequency, where #alpha=(Δomega)/( Δt)#. We can find #omega# from the given change in frequency of #15Hz#, as #omega=2pif#.
#omega=2pi(15s^-1)#
#omega=30pi(rad)/s#
Because the problem states that the frequency changed by this amount, this is # Δomega#. We are given that this took place over #6s#, which is our # Δt# value. Thus,
#alpha=(30pi(rad)/s)/(6s)#
#alpha=5pi(rad)/s^2#
Now that we have values for #I# and #alpha#, we can calculate the torque.
#tau=Ialpha#
#tau=50/12kgm^2*5pi (rad)/s^2#
#tau=(250pi)/12Nm#
#tau≈65Nm#
