# An object with a mass of  2 kg is traveling in a circular path of a radius of 2 m. If the object's angular velocity changes from  3 Hz to  8 Hz in  1 s, what torque was applied to the object?

Mar 28, 2018

The torque is $= 251.3 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

where $I$ is the moment of inertia

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

The mass is $m = 2 k g$

The radius of the path is $r = 2 m$

So the moment of inertia is

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

The angular acceleration is

$\alpha = \frac{2 \pi {f}_{1} - 2 \pi {f}_{0}}{t} = \left(16 - 6\right) \frac{\pi}{1} = 10 \pi r a {\mathrm{ds}}^{-} 2$

So,

The torque is

$\tau = I \alpha = 8 \cdot 10 \pi = 251.3 m$