# An object with a mass of  2 kg is traveling in a circular path of a radius of 4 m. If the object's angular velocity changes from  2 Hz to  9 Hz in  2 s, what torque was applied to the object?

Jan 16, 2018

The torque is $= 703.7 N m$

#### 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 $m = 2 k g$

The radius of the path is $r = 4 m$

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

So, $I = 2 \cdot {\left(4\right)}^{2} = 32 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{{f}_{2} - {f}_{1}}{t} \cdot 2 \pi = \frac{9 - 2}{2} \cdot 2 \pi$

$= \left(7 \pi\right) r a {\mathrm{ds}}^{- 2}$

So,

The torque is $\tau = 32 \cdot 7 \pi = 703.7 N m$