# 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 1 s?

Jan 3, 2017

The torque is $= 50.3 N m$

#### Explanation:

We asume that the rod is spinning around its centerm this is important for the moment of inertia.

The torque is the rate of change of angular momentum

$\tau = \frac{\mathrm{dL}}{\mathrm{dt}} = \frac{d \left(I \omega\right)}{\mathrm{dt}} = I \frac{\mathrm{do} m e g a}{\mathrm{dt}}$

The moment of inertia of a rod is $I = \frac{1}{12} \cdot m {L}^{2}$

$= \frac{1}{12} \cdot 3 \cdot {4}^{2} = 4 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{2}{1} \cdot 2 \pi$

$= \left(4 \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 4 \cdot 4 \pi N m = 16 \pi N m = 50.3 N m$