# 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?

May 31, 2016

I got $5. \overline{33} \pi$ $\text{N"cdot"m}$.

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

$\setminus m a t h b f \left(\sum \vec{\tau} = I \overline{\alpha}\right)$

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 $\overline{\alpha}$. For this, we have:

$\setminus m a t h b f \left(\overline{\alpha} = \frac{\Delta \vec{\omega}}{\Delta t} = \frac{{\vec{\omega}}_{f} - {\vec{\omega}}_{i}}{{t}_{f} - {t}_{i}} = \frac{{\vec{\omega}}_{f}}{t}\right) ,$

where if we assume the initial angular velocity ${\vec{\omega}}_{i} = 0$, then ${\vec{\omega}}_{f} = \Delta \vec{\omega}$, and with initial time ${t}_{i} = 0 , {t}_{f} = t$.

Next, for this ideal rod of length $L$, to find its moment of inertia ${I}_{\text{cm}}$ about its center of mass, we need to take the integral for a uniform rod rotating about its center of mass (the midpoint $\left(0 , 0\right)$), with infinitesimal mass $\mathrm{dm} = \frac{m}{L} \mathrm{dr}$:

$\textcolor{g r e e n}{{I}_{\text{cm}}} = \frac{m}{L} {\int}_{- \frac{L}{2}}^{\frac{L}{2}} {r}^{2} \mathrm{dr}$

where $r$ would be the distance from the center of mass to the end of the rod, or $\frac{L}{2}$.

$= \frac{m}{L} | \left[{r}^{3} / 3\right] {|}_{- \frac{L}{2}}^{\frac{L}{2}}$

$= \frac{m}{L} \left[{\left(\frac{L}{2}\right)}^{3} / 3 - {\left(- \frac{L}{2}\right)}^{3} / 3\right]$

$= \frac{m}{L} \left[{L}^{3} / 24 + {L}^{3} / 24\right]$

$= \frac{m}{L} \left[{L}^{3} / 12\right]$

$= \textcolor{g r e e n}{\frac{1}{12} m {L}^{2}}$

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

$2 \pi f = \omega$

So, the torque needed should be:

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

$= \frac{m {L}^{2}}{12} \frac{{\vec{\omega}}_{f}}{t}$

$= \left(\left(\text{3 kg"cdot("4 m")^2)/12)((2pi*"2 s"^(-1))/("3 s}\right)\right)$

$= \textcolor{b l u e}{5. \overline{\text{33}} \pi}$ $\textcolor{b l u e}{\text{N"cdot"m}}$