# An object with a mass of  3 kg is traveling in a circular path of a radius of 2 m. If the object's angular velocity changes from  1 Hz to  3 Hz in  5 s, what torque was applied to the object?

Jan 4, 2016

$= 60.75 N . m$.

#### Explanation:

From rotational dynamics we get that $\sum \vec{\tau} = I \vec{\alpha}$, where $\tau$ is the torque, $I$ is the moment of inertia, and $\alpha$ is the angular acceleration.

But since angular acceleration is the rate of change in angular velocity, we get that $\alpha = \frac{\mathrm{do} m e g a}{\mathrm{dt}}$.

We then first need to convert the linear frequency in Hertz into angular frequency in rad/s.

$\alpha = \frac{\mathrm{do} m e g a}{\mathrm{dt}} = \frac{\left(3 \times 2 \pi r\right) - \left(1 \times 2 \pi r\right)}{5} = 5.0265 r a d / {s}^{2}$.

Moment of inertia $I = {\sum}_{j} {m}_{j} {r}_{p e r p j}^{2} = {\int}_{m} {r}^{2} \mathrm{dm}$.

So in this case for a point mass, $I = m {r}^{2} = 3 \times {2}^{2} = 12 k g . {m}^{2}$.

Therefore resultant torque is $\tau = I \alpha$

$= 12 \times 5.0265$

$= 60.75 N . m$.