# An object with a mass of  8 kg is traveling in a circular path of a radius of 12 m. If the object's angular velocity changes from  9 Hz to  12 Hz in  6 s, what torque was applied to the object?

Apr 13, 2017

The torque was $= 3619.1 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

For the object, the moment of inertia is

$I = \left(m {r}^{2}\right)$

So, $I = 8 \cdot {\left(12\right)}^{2} = 1152 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{12 - 9}{6} \cdot 2 \pi$

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

So the torque is $\tau = 1152 \cdot \left(\pi\right) = 3619.1 N m$