# What torque would have to be applied to a rod with a length of 8 m and a mass of 8 kg to change its horizontal spin by a frequency 1 Hz over 9 s?

Feb 23, 2017

The torque (for the rod rotating about the center) is $= 29.8 N m$
The torque (for the rod rotating about one end) is $= 119.1 N m$

#### Explanation:

The moment of inertia of a rod, rotating about the center is

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

$= \frac{1}{12} \cdot 8 \cdot {8}^{2} = \frac{128}{3} k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{1}{9} \cdot 2 \pi$

$= \left(\frac{2}{9} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = \frac{128}{3} \cdot \left(\frac{2}{9} \pi\right) N m = \frac{256}{27} \pi N m = 29.8 N m$

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

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

$= \frac{1}{3} \cdot 8 \cdot {8}^{2} = \frac{512}{3} k g {m}^{2}$

So,

The torque is $\tau = \frac{512}{3} \cdot \left(\frac{2}{9} \pi\right) = \frac{1024}{27} \pi = 119.1 N m$