What torque would have to be applied to a rod with a length of 4 m4m and a mass of 3 kg3kg to change its horizontal spin by a frequency of 5 Hz5Hz over 1 s1s?

1 Answer
Dec 16, 2016

tau≈126Nmτ126Nm

Explanation:

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=1123kg(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