# What torque would have to be applied to a rod with a length of 3 m and a mass of 1 kg to change its horizontal spin by a frequency of 5 Hz over 2 s?

Mar 30, 2018

The torque for the rod rotating about the center is $= 11.78 N m$
The torque for the rod rotating one end is $= 47.12 N m$

#### Explanation:

The torque is the rate of change of angular momentum

$\tau = \frac{\mathrm{dL}}{\mathrm{dt}} = \frac{d \left(I \omega\right)}{\mathrm{dt}} = I \frac{\mathrm{do} m e g a}{\mathrm{dt}} = I \alpha$

The mass of the rod is $m = 1 k g$

The length of the rod is $L = 3 m$

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

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

$= \frac{1}{12} \cdot 1 \cdot {3}^{2} = 0.75 k g {m}^{2}$

The rate of change of angular velocity is

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

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

So the torque is $\tau = 0.75 \cdot \left(5 \pi\right) N m = 11.78 N m$

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

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

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

So,

The torque is

$= 3 \cdot \left(5 \pi\right) = 47.12 N m$