# What torque would have to be applied to a rod with a length of 3 m and a mass of 8 kg to change its horizontal spin by a frequency 8 Hz over 5 s?

Jun 5, 2017

The torque for the rod rotating about the center is $= 60.3 N m$
The torque for the rod rotating about one end is $= 241.3 N m$

#### 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

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

$= \frac{1}{12} \cdot 8 \cdot {3}^{2} = 6 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{8}{5} \cdot 2 \pi$

$= \left(\frac{16}{5} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 6 \cdot \left(\frac{16}{5} \pi\right) N m = \frac{96}{5} \pi N m = 60.3 N m$

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

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

$= \frac{1}{3} \cdot 8 \cdot {3}^{2} = 24 k g {m}^{2}$

So,

The torque is $\tau = 24 \cdot \left(\frac{16}{5} \pi\right) = \frac{3844}{5} \pi = 241.3 N m$