# An object with a mass of  3 kg is traveling in a circular path of a radius of 5 m. If the object's angular velocity changes from  7 Hz to  8 Hz in  5 s, what torque was applied to the object?

Nov 16, 2017

The torque is $= 157.1 N m$

#### 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

The mass of the object is $m = 3 k g$

The radius of the path is $r = 5 m$

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

So, $I = 3 \cdot {\left(5\right)}^{2} = 75 k g {m}^{2}$

And the rate of change of angular velocity is

$\frac{d \omega}{\mathrm{dt}} = \frac{\Delta \omega}{t} = \frac{2 \pi {f}_{1} - 2 \pi {f}_{2}}{t}$

$= \frac{16 \pi - 14 \pi}{5} = \left(\frac{2}{5} \pi\right) r a {\mathrm{ds}}^{-} 2$

So,

The torque is

$\tau = 75 \cdot \frac{2}{5} \pi = 157.1 N m$