An object with a mass of  7 kg is traveling in a circular path of a radius of 9 m. If the object's angular velocity changes from  5 Hz to  3 Hz in  8 s, what torque was applied to the object?

Jun 26, 2017

The torque was $= 890.6 N m$

Explanation:

The torque is the rate of change of angular momentum

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

where $I$ is the moment of inertia

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

So, $I = 7 \cdot {\left(9\right)}^{2} = 567 k g {m}^{2}$

The rate of change of angular velocity is

$\frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{5 - 3}{8} \cdot 2 \pi$

$= \left(\frac{1}{2} \pi\right) r a {\mathrm{ds}}^{- 2}$

So the torque is $\tau = 567 \cdot \left(\frac{1}{2} \pi\right) N m = 890.6 N m$