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

Dec 25, 2016

The torque is $= 294.5 N m$

#### Explanation:

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 (spinning around the middle of the rod) is $I = \frac{1}{12} m {l}^{2}$

$= \frac{1}{2} \cdot 5 \cdot {5}^{2} = \frac{125}{2} k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{3}{4} \cdot 2 \pi$

$= \frac{3}{2} \pi r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = \frac{125}{2} \cdot \frac{3}{2} \pi N m = \frac{375}{4} \pi N m = 294.5 N m$