# What torque would have to be applied to a rod with a length of 6 m and a mass of 3 kg to change its horizontal spin by a frequency 5 Hz over 4 s?

Sep 3, 2016

$\tau = 657.75$ $N - m$

#### Explanation:

The torque is given by:
$\tau = I \cdot \dot{\omega} = I \cdot \frac{\omega}{t}$
where $I$ is the moment of Inertia of the horizontal spinning rod which is given by: $I = \frac{m \cdot {l}^{2}}{12}$
and $\omega$, the angular velocity is given by $\omega = 2 \pi \cdot f$

So, from the given data
$I = \frac{3 \cdot {6}^{2}}{12} = 9$ $k g - {m}^{2}$
$\omega = 2 \pi \cdot 5 = 31.4$ $\frac{r a d}{s}$
$t = 4$ $s$

Thus we can now evaluate the torque
$\tau = I \cdot \frac{\omega}{t}$
$\tau = 9 \cdot \frac{31.4}{4} = 657.75$ $N - m$