# 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  8 Hz in  5 s, what torque was applied to the object?

Dec 26, 2015

It depends on whether it's a vertical loop or a horizontal loop, but let's assume it's horizontal and imagine a so-called "massless" string attached to the object.

Let's try drawing a Free-Body diagram for this: Now, we can use a simple kinematics equation analog relating $\Delta \vec{\omega}$ with $\vec{\alpha}$.

Just like how:

${\vec{v}}_{f} = {\vec{v}}_{i} + \vec{a} t$

... we have:

$\setminus m a t h b f \left({\vec{\omega}}_{f} = {\vec{\omega}}_{i} + \vec{\alpha} t\right)$

So with this, we can find $\alpha$, and with $\alpha$, we can input that into this equation for the sum of the torques:

$\setminus m a t h b f \left(\sum \tau = I \alpha\right)$

Next, we need the inertia for a mass rotating about an axis. It should be $I = k m {r}^{2}$, where $k = 1$. So:

$\sum {\tau}_{\text{ext}}$

$= \textcolor{g r e e n}{I} \textcolor{h i g h l i g h t}{\alpha}$

$= \textcolor{g r e e n}{m {r}^{2}} \textcolor{h i g h l i g h t}{\left(\frac{{\vec{\omega}}_{f} - {\vec{\omega}}_{i}}{t}\right)}$

$= \textcolor{g r e e n}{\left(\text{3 kg")("2 m")^2)color(highlight)((("8 rad/s" - "1 rad/s")/"5 s}\right)}$

$= \textcolor{b l u e}{\text{16.8 N"cdot"m}}$