# What torque would have to be applied to a rod with a length of 7 m and a mass of 12 kg to change its horizontal spin by a frequency 15 Hz over 9 s?

Dec 26, 2017

The torque for the rod rotating about the center is $= 513.1 N m$
The torque for the rod rotating about one end is $= 2052.5 N m$

#### 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 mass of the rod is $m = 12 k g$

The length of the rod is $L = 7 m$

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

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

$= \frac{1}{12} \cdot 12 \cdot {7}^{2} = 49 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{15}{9} \cdot 2 \pi$

$= \left(\frac{10}{3} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 49 \cdot \left(\frac{10}{3} \pi\right) N m = 513.1 N m$

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

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

$= \frac{1}{3} \cdot 12 \cdot {7}^{2} = 196 k g {m}^{2}$

So,

The torque is $\tau = 196 \cdot \left(\frac{10}{3} \pi\right) = 2052.5 N m$