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

Feb 14, 2018

The torque is $= 402.1 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

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

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

The initial angular velocity is

${\omega}_{0} = 2 \pi {f}_{0} = 2 \cdot \pi \cdot 1 = 2 \pi r a {\mathrm{ds}}^{-} 1$

The final angular velocity is

${\omega}_{1} = 2 \pi {f}_{1} = 2 \cdot \pi \cdot 5 = 10 \pi r a {\mathrm{ds}}^{-} 1$

The time is $t = 2 s$

The angular acceleration is

$\alpha = \frac{{\omega}_{1} - {\omega}_{0}}{t} = \frac{10 \pi - 2 \pi}{2} = 4 \pi r a {\mathrm{ds}}^{-} 2$

So,

The torque is $\tau = 32 \cdot 4 \pi = 402.1 N m$