The net torque, tauτ, can be expressed in terms of the moment of inertia and angular acceleration, where tau=Ialphaτ=Iα.
Assuming that this is a thin, rigid rod, with the axis of rotation through its center, the moment of inertia, II, of the rod is given by 1/12ML^2112ML2, where MM is the mass of the rod and LL is its length. If the rod is rotated about its end, I=1/3ML^2I=13ML2. I will show the calculation for the axis of rotation through the center. We are given both LL and MM. Thus,
I=1/12*3kg*(4m)^2I=112⋅3kg⋅(4m)2
I=4kgm^2I=4kgm2
The angular acceleration of the rod as it rotates, alphaα, can be calculated from the given values of time and frequency, where alpha=(Δomega)/( Δt). We can find omega from the given frequency of 5Hz, as omega=2pif.
omega=2pi(5s^-1)
omega=10pi(rad)/s
Because the problem states that the frequency changed by this amount, this is Δomega. We are given that this took place over 1s, which is our Δt value. Thus,
alpha=(10pi(rad/s))/(1s)
alpha=10pi(rad)/s^2
Now that we have all of the values we need, we can calculate the torque.
tau=Ialpha
tau=4kgm^2*10pi(rad)/s^2
tau=40piNm
tau≈126Nm