# What torque would have to be applied to a rod with a length of 7 m and a mass of 9 kg to change its horizontal spin by a frequency 9 Hz over 18 s?

Apr 5, 2017

The torque, for the rod rotating about the center is, $= 115.5 N m$
The torque, for the rod rotating about one end is, $= 461.8 N m$

#### 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

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

$= \frac{1}{12} \cdot 9 \cdot {7}^{2} = 36.75 k g {m}^{2}$

The rate of change of angular velocity is

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

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

So the torque is $\tau = 36.75 \cdot \left(\pi\right) N m = 36.75 \pi N m = 115.5 N m$

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

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

$= \frac{1}{3} \cdot 9 \cdot {7}^{2} = 147 k g {m}^{2}$

So,

The torque is $\tau = 147 \cdot \left(\pi\right) = 461.8 N m$

Apr 5, 2017

$\text{Torque: "36.75pi" } N . m$

#### Explanation:

$\text{Let's apply Newton's second law for circular motion.}$

$F = m \cdot a \text{ For linear motion}$

$T = I \cdot \alpha \text{ For circular motion}$

$\text{Where T:Torque , I:Moment of Inertia , "alpha:"angular acceleration}$

I=1/12 m* l^2"
$\text{moment of inertia for a rod turning about its mass center}$
$m = 9 k g \text{ , } l = 7 m$

$I = \frac{1}{12} \cdot 9 \cdot {7}^{2} = 36.75$

$\alpha = \frac{\Delta \omega}{\Delta t}$

$\Delta \omega = {\omega}_{2} - {\omega}_{1}$

$\Delta \omega = 2 \pi \left({f}_{2} - {f}_{1}\right)$

${f}_{2} - {f}_{1} = 9 H z$

$\Delta \omega = 18 \pi$

$\alpha = \frac{18 \pi}{18}$

$\alpha = \pi$

$T = 36.75 \pi \text{ } N \cdot m$