# What torque would have to be applied to a rod with a length of 4 m and a mass of 2 kg to change its horizontal spin by a frequency 8 Hz over 8 s?

Feb 9, 2016
1. If the rod spins about an axis passing through its center:
$\setminus \tau = I \setminus \frac{d \setminus \omega}{\mathrm{dt}} = \left(\frac{8}{3} k g . {m}^{2}\right) \left(2 \setminus \pi \frac{r a d}{s}\right) = \setminus \frac{16 \setminus \pi}{3} N . m$
2. If the rod spins about an axis passing through its end:$\setminus \tau = I \setminus \frac{d \setminus \omega}{\mathrm{dt}} = \left(\frac{32}{3} k g . {m}^{2}\right) \left(2 \setminus \pi \frac{r a d}{s}\right) = \setminus \frac{64 \setminus \pi}{3} N . m$

#### Explanation:

Torque: $\setminus \tau \setminus \equiv \setminus \frac{\mathrm{dL}}{\mathrm{dt}} = I \setminus \alpha = I \setminus \frac{d \setminus \omega}{\mathrm{dt}}$

$L$ - Angular Momentum; $\setminus \quad$ $I$ - Moment-of-inertia;
$\alpha$ - Angular acceleration; $\setminus \quad$ $\setminus \omega$ - Angular frequency;
$f$ - frequency.

\omega = 2\pi f; \qquad \frac{d\omega}{dt} = 2\pi\frac{df}{dt}=2\pi\frac{8Hz}{8s}=2\pi $\frac{r a d}{s} ^ 2$.

The moment-of-inertia $I$ of the rod depends on the location of the spin axis. If $M$ is the mass of the rod and $L$ its length then the moment-of-inertia of the rod is:

$M = 2$ kg $\setminus q \quad$ $L = 4$ m

1. If the rod spins about an axis passing through its centre,
$I = \setminus \frac{1}{12} M {L}^{2} = \setminus \frac{1}{12} \left(2 k g\right) {\left(4 m\right)}^{2} = \frac{8}{3} k g . {m}^{2}$
$\setminus \tau = I \setminus \frac{d \setminus \omega}{\mathrm{dt}} = \left(\frac{8}{3} k g . {m}^{2}\right) \left(2 \setminus \pi \frac{r a d}{s} ^ 2\right) = \setminus \frac{16 \setminus \pi}{3} N . m$
2. If the rod spins about an axis passing through its end,
$I = \setminus \frac{1}{3} M {L}^{2} = \setminus \frac{1}{3} \left(2 k g\right) {\left(4 m\right)}^{2} = \setminus \frac{32}{3} k g . {m}^{2}$
$\setminus \tau = I \setminus \frac{d \setminus \omega}{\mathrm{dt}} = \left(\frac{32}{3} k g . {m}^{2}\right) \left(2 \setminus \pi \frac{r a d}{s} ^ 2\right) = \setminus \frac{64 \setminus \pi}{3} N . m$