# What torque would have to be applied to a rod with a length of 7 m and a mass of 9 kg to change its horizontal spin by a frequency 12 Hz over 16 s?

Feb 21, 2017

The torque (about the center) is $= 173.2 N m$
The torque (about one end) is $= 692.7 N m$

#### 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 abour the center is

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

$= \frac{1}{12} \cdot 9 \cdot {7}^{2} = \frac{147}{4} k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{12}{16} \cdot 2 \pi$

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

So the torque is $\tau = \frac{147}{4} \cdot \frac{3 \pi}{2} N m = \frac{441}{8} \pi N m = 173.2 N m$

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

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

$= \frac{1}{3} \cdot 9 \cdot {7}^{2} = 147$

So,

The torque is $\tau = 147 \cdot \left(\frac{3}{2} \pi\right) = \frac{441}{2} \pi = 692.7 N m$