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

##### 1 Answer
Sep 14, 2017

We will need to know the angular acceleration that existed during those 3 s. If this was a linear motion problem we could use the suvat formula
$v = u + a \cdot t$
The equivalent formula in angular motion is
${\omega}_{f} = {\omega}_{i} + \alpha \cdot t$

Using that last formula,
$5 \cdot \frac{2 \cdot \pi \text{ radians")/s = 4*(2*pi " radians}}{s} + \alpha \cdot 3 s$

Solving for $\alpha$,
alpha = ((2*pi " radians")/s)/(3 s) = 2.09 " radians"/s^2

Next we need the angular motion equivalent of F=m*a (Newton's 2nd Law). The equivalent is

$\tau = I \cdot \alpha$

I will assume the object can be considered a point mass. This requires that the dimensions of the object be much smaller than the 4 m radius. With that assumption, the rotational inertia, $I$, of the rotating mass is
$I = m \cdot {r}^{2} = 2 k g \cdot {\left(4 m\right)}^{2} = 32 k g \cdot {m}^{2}$

So now we can calculate the torque, $\tau$.

$\tau = 32 k g \cdot {m}^{2} \cdot 2.09 \frac{\text{ radians}}{s} ^ 2 = 67 k g \cdot {m}^{2} / {s}^{2}$
$\tau = 67 N \cdot m$

You might ask how the combination of units $k g \cdot {m}^{2} / {s}^{2}$ simplified into $N \cdot m$. Note that you can group $k g \cdot \frac{m}{s} ^ 2$ apart from the other m. And $k g \cdot \frac{m}{s} ^ 2$ is the definition of the Newton.

Torque is the product of force and the length of the lever arm. So, we have our answer.

I hope this helps,
Steve