# What torque would have to be applied to a rod with a length of 8 m and a mass of 3 kg to change its horizontal spin by a frequency 18 Hz over 9 s?

Mar 2, 2017

The torque (for the rod rotating about the center) is $= 201.1 N m$
The torque (for the rod rotating about one end) is $= 804.2 N m$

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

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

$= \frac{1}{12} \cdot 3 \cdot {8}^{2} = 16 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{18}{9} \cdot 2 \pi$

$= \left(4 \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 16 \cdot \left(4 \pi\right) N m = 64 \pi N m = 201.1 N m$

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

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

$= \frac{1}{3} \cdot 3 \cdot {8}^{2} = 64 k g {m}^{2}$

So,

The torque is $\tau = 64 \cdot \left(4 \pi\right) = 804.2 N m$