# An object with a mass of  7 kg is traveling in a circular path of a radius of 9 m. If the object's angular velocity changes from  5 Hz to  6 Hz in  8 s, what torque was applied to the object?

Jan 13, 2018

442.26 Nm

#### Explanation:

Torque acting on a particle is given as, $\tau$ = I$\propto$
(Where, I is the moment if inertia and $\propto$ is angular acceleration)
Now,$I = m \left({r}^{2}\right)$
Or, I= 567 SI unit
$\propto$ = $\frac{\mathrm{do} m e g a}{\mathrm{dt}}$
= ($\omega$ final -$\omega$ initial)/$t$ where $\omega$ is angular velocity
= $\left(2 \frac{\pi}{t}\right)$ (n final - n initial) (as omega=2$\pi$n) where n is frequency
= 0.78 $\frac{r a d}{{s}^{2}}$
Hence, $\tau$ = $\left(567 \cdot 0.78\right)$N.m
Or,442.26 N.m

Jan 13, 2018

The torque is $= 445.3 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

where $I$ is the moment of inertia

The mass of the object is $m = 7 k g$

The radius of the path is $r = 9 m$

For the object, the moment of inertia is $I = m {r}^{2}$

So, $I = 7 \cdot {\left(9\right)}^{2} = 567 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{{f}_{2} - {f}_{1}}{t} \cdot 2 \pi = \frac{6 - 5}{8} \cdot 2 \pi$

$= \left(\frac{1}{4} \pi\right) r a {\mathrm{ds}}^{- 2}$

So,

The torque is $\tau = 567 \cdot \frac{1}{4} \pi = 445.3 N m$