Question

What torque would have to be applied to a rod with a length of `5 m` and a mass of `2 kg` to change its horizontal spin by a frequency of `5 Hz` over `4 s`?

Answer 1

`tau≈33Nm`

Explanation:

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`

`=25/6kgm^2`

The angular acceleration of the rod as it rotates, `alpha`, can be calculated from the given values of time and frequency, given by:

`alpha=(Δomega)/( Δt)`.

We can find `omega` from the given change in frequency of `5Hz`, as `omega=2pif`.

`omega=2pi(5s^-1)`

`=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 `4s`, which is our `Δt` value. Thus,

`alpha=(10pi(rad)/s)/(4s)`

`=(5pi)/2(rad)/s^2`

Now that we have values for `I` and `alpha`, we can calculate the torque:

`tau=Ialpha`

`=25/6kgm^2*(5pi)/2(rad)/s^2`

`=(125pi)/12Nm`

`≈33Nm`