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 9 Hz over 6 s?

Feb 21, 2016

$L = \frac{25 \cdot \pi}{2}$

Explanation:

$L = \alpha \cdot I$
$\text{(L:torque ;alpha :angular acceleration ;I:the moment of inertia)}$
$\alpha = \frac{\Delta \omega}{\Delta t}$
$\alpha = \frac{2 \cdot \pi \cdot \Delta f}{\Delta t}$
$\alpha = \frac{2 \cdot \pi \cdot 9}{6}$
$\alpha = 3 \cdot \pi$
$I = \frac{1}{12} \cdot m \cdot {l}^{2} \text{ for a rod spinning around its center}$
$I = \frac{1}{12} \cdot 2 \cdot {5}^{2} = \frac{25}{6}$
$L = 3 \cdot \pi \cdot \frac{25}{6}$
$L = \frac{25 \cdot \pi}{2}$