# 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  4 Hz to  9 Hz in  3 s, what torque was applied to the object?

The torque applied to the object is 53.33 Newton-metre.

#### Explanation:

As we know torque $\tau = I \alpha$
So angular acceleration $\alpha$= (change in angular velocity omega)//(change in time t)
$\implies \alpha = \left({\omega}_{2} - {\omega}_{1}\right) / \mathrm{dt}$
$\implies \alpha = 5 / 3 \frac{m}{s} ^ 2$
And torque now becomes $\tau = m {r}^{2} \alpha \left[I = m {r}^{2}\right]$
By substituting all values we get,
$\tau = 2 k g \times {\left(4 m\right)}^{2} \times \left(\frac{5}{3} \frac{m}{s} ^ 2\right) = 53.33$Newton-metre.