# An object with a mass of  3 kg is traveling in a circular path of a radius of 7 m. If the object's angular velocity changes from  14 Hz to  23 Hz in 9 s, what torque was applied to the object?

Apr 5, 2017

The torque was $= 923.6 N m$

#### Explanation:

The torque is the rate of change of angular momentum

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

The moment of inertia of a rod, rotating about the center is

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

$= 3 \cdot {7}^{2} = 147 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{23 - 14}{9} \cdot 2 \pi$

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

So the torque is $\tau = 147 \cdot \left(2 \pi\right) N m = 923.6 N m$