# 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  12 Hz in  2 s, what torque was applied to the object?

Jul 4, 2018

The torque is $= 1005.3 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

For the object, the moment of inertia is

$I = m {r}^{2}$

The mass of the object is $m = 2 k g$

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

The moment of inertia is

$I = 2 \cdot {4}^{2} = 32 k g {m}^{2}$

The angular acceleration is

$\alpha = 2 \pi \frac{{f}_{2} - {f}_{1}}{t} = 2 \pi \frac{12 - 2}{2} = 10 \pi r a {\mathrm{ds}}^{-} 2$

The torque is

$\tau = I \omega = 32 \cdot 10 \pi = 1005.3 N m$