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  13Hz to  17 Hz in 5 s, what torque was applied to the object?

Jun 27, 2018

The torque was $= 738.9 N m$

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 disc is $m = 3 k g$

The radius of the disc is $r = 7 m$

The moment of inertia is

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

The angular acceleration is

$\alpha = 2 \pi \frac{{f}_{2} - {f}_{1}}{t} = 2 \pi \frac{17 - 13}{5} = 1.6 \pi r a {\mathrm{ds}}^{-} 2$

The torque is

$\tau = I \omega = 147 \cdot 1.6 \pi = 738.9 N m$