What torque would have to be applied to a rod with a length of `4 m` and a mass of `3 kg` to change its horizontal spin by a frequency of `2 Hz` over `3 s`?

I got `5.bar(33)pi` `"N"cdot"m"`.

The sum of the torques for a rod is written as:

`\mathbf(sum vectau = Ibaralpha)`

I assume we are looking at a rod rotating horizontally on a flat surface (so that no vertical forces apply), changing in angular velocity over time.

Then, we are looking at the average angular acceleration `baralpha`. For this, we have:

`\mathbf(baralpha = (Deltavecomega)/(Deltat) = (vecomega_f - vecomega_i)/(t_f - t_i) = (vecomega_f)/t),`

where if we assume the initial angular velocity `vecomega_i = 0`, then `vecomega_f = Deltavecomega`, and with initial time `t_i = 0, t_f = t`.

Next, for this ideal rod of length `L`, to find its moment of inertia `I_"cm"` about its center of mass, we need to take the integral for a uniform rod rotating about its center of mass (the midpoint `(0,0)`), with infinitesimal mass `dm = m/Ldr`:

`color(green)(I_"cm") = m/Lint_(-L/2)^(L/2) r^2dr`

where `r` would be the distance from the center of mass to the end of the rod, or `L/2`.

`= m/L|[r^3/3]|_(-L/2)^(L/2)`

`= m/L[(L/2)^3/3 - (-L/2)^3/3]`

`= m/L[L^3/24 + L^3/24]`

`= m/L[L^3/12]`

`= color(green)(1/12mL^2)`

Lastly, we should recognize that the frequency is not equal to `omega`. Actually:

`2pif = omega`

So, the torque needed should be:

`color(blue)(sum vectau) = vectau_(vec"F"_"app") = I_"cm"baralpha`

`= (mL^2)/12(vecomega_f)/t`

`= (("3 kg"cdot("4 m")^2)/12)((2pi*"2 s"^(-1))/("3 s"))`

`= color(blue)(5.bar"33"pi)` `color(blue)("N"cdot"m")`