# What torque would have to be applied to a rod with a length of 4 m and a mass of 3 kg to change its horizontal spin by a frequency of 5 Hz over 2 s?

Dec 5, 2017

The torque for the rod rotating about its center is $= 62.83 N m$
The torque for the rod rotating about one end is $= 251.33 N m$

#### 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 $m = 3 k g$

The length of the rod is $L = 4 m$

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

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

$= \frac{1}{12} \cdot 3 \cdot {4}^{2} = 4 k g {m}^{2}$

The frequency is $f = 5 H z$

The rate of change of angular velocity is

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

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

So the torque is $\tau = 4 \cdot \left(5 \pi\right) N m = 20 \pi N m = 62.83 N m$

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

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

$= \frac{1}{3} \cdot 3 \cdot {4}^{2} = 16 k g {m}^{2}$

So,

The torque is $\tau = 16 \cdot \left(5 \pi\right) = 80 \pi = 251.33 N m$