# What torque would have to be applied to a rod with a length of 1 m and a mass of 2 kg to change its horizontal spin by a frequency of 6 Hz over 4 s?

Feb 10, 2016

The angular acceleration is $\left(3 \pi\right) \textrm{r a \mathrm{di} a n s} \cdot {s}^{-} 2$ and the moment of inertia is 2 kg ${m}^{2}$. The required torque is the product of these, which is 6$\pi$ Nm, or approximately 18.8 Nm

#### Explanation:

First find the angular acceleration, $\alpha$.

$\alpha = \frac{\textrm{c h a n \ge \in a n g \underline{a} r v e l o c i t y}}{\textrm{t i m e}}$

The angular velocity is 6 Hz, and 1 Hz means that a circle of 2$\pi$ radians is traversed in 1 second. Thus the angular velocity is 6 (2 $\pi$ radians) ${s}^{-} 1$, which is 12 $\pi$ radians ${s}^{-} 1$

$\alpha$ = 6 (2 $\pi$ *radians) ${s}^{-} 1$) / 4 s = (3 $\pi$ * radians) ${s}^{-} 2$

Torque = moment of inertia * $\alpha$

moment of inertia depends on the shape of the spinning object and can be complicated. Fortunately for a mass on a rod it's simple:
moment of inertia = mass * ${\textrm{r o d \le n > h}}^{2}$
so:
moment of inertia = 2kg * $1 {m}^{2}$ = 2 kg ${m}^{2}$

Now we have:

Torque = (2 kg ${m}^{2}$) * ((3 pi) text{radians} (s^-2))

Torque = 6$\pi$ Nm,
or approximately :
Torque ~ 18.8 Nm