# 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  1 Hz to  5 Hz in  3 s, what torque was applied to the object?

Apr 7, 2017

The torque is $= 268.1 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

Where $I$ is the moment of inertia

For the object, $I = m {r}^{2}$

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

The rate of change of angular velocity is

$\frac{\Delta \omega}{\Delta t} = \frac{\left(5 - 1\right) \cdot 2 \pi}{3} = \frac{8}{3} \pi r a {\mathrm{ds}}^{-} 2$

So,

The torque is

$\tau = 32 \cdot \frac{8}{3} \pi = 268.1 N m$