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
