# An object with a mass of  3 kg is traveling in a circular path of a radius of 7 m. If the object's angular velocity changes from  6 Hz to  8 Hz in  8 s, what torque was applied to the object?

Jun 7, 2017

The torque was $= 230.9 N m$

#### 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 the moment of inertia is $= I$

and $\omega$ is the angular velocity

The mass is $m = 3 k g$

The radius is $r = 7 m$

For the object, $I = \left(m {r}^{2}\right)$

So, $I = 3 \cdot {\left(7\right)}^{2} = 147 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{8 - 6}{8} \cdot 2 \pi$

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

So the torque is $\tau = 147 \cdot \left(\frac{1}{2} \pi\right) N m = \frac{147}{2} \pi N m = 230.9 N m$