# 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 5 Hz over 6 s?

Mar 23, 2017

The torque, for the rod rotating about the center is $= 21.82 N m$
The torque, for the rod rotating about one end is $= 87.28 N m$

#### 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 of a rod, rotating about the center is

$I = \frac{1}{12} \cdot m {L}^{2}$

$= \frac{1}{12} \cdot 2 \cdot {5}^{2} = \frac{25}{6} k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{5}{6} \cdot 2 \pi$

$= \left(\frac{5}{3} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = \frac{25}{6} \cdot \left(\frac{5}{3} \pi\right) N m = \frac{125}{18} \pi N m = 21.82 N m$

The moment of inertia of a rod, rotating about one end is

$I = \frac{1}{3} \cdot m {L}^{2}$

$= \frac{1}{3} \cdot 2 \cdot {5}^{2} = \frac{50}{3}$

So,

The torque is $\tau = \frac{50}{3} \left(\frac{5}{3} \pi\right) = \frac{250}{9} \pi = 87.28 N m$