# What torque would have to be applied to a rod with a length of 5 m and a mass of 2 kg to change its horizontal spin by a frequency of 15 Hz over 6 s?

Dec 28, 2016

tau≈65Nm

#### Explanation:

The net torque, $\tau$, can be expressed in terms of the moment of inertia and angular acceleration, where $\tau = I \alpha$.

Assuming that this is a thin, rigid rod, with the axis of rotation through its center, the moment of inertia, $I$, of the rod is given by $\frac{1}{12} M {L}^{2}$, where $M$ is the mass of the rod and $L$ is its length. If the rod is rotated about its end, $I = \frac{1}{3} M {L}^{2}$. I will show the calculation for the axis of rotation through the center. We are given both $L$ and $M$. Thus,

$I = \frac{1}{12} \cdot 2 k g \cdot {\left(5 m\right)}^{2}$

$I = \frac{50}{12} k g {m}^{2}$

The angular acceleration of the rod as it rotates, $\alpha$, can be calculated from the given values of time and frequency, where alpha=(Δomega)/( Δt). We can find $\omega$ from the given change in frequency of $15 H z$, as $\omega = 2 \pi f$.

$\omega = 2 \pi \left(15 {s}^{-} 1\right)$

$\omega = 30 \pi \frac{r a d}{s}$

Because the problem states that the frequency changed by this amount, this is  Δomega. We are given that this took place over $6 s$, which is our  Δt value. Thus,

$\alpha = \frac{30 \pi \frac{r a d}{s}}{6 s}$

$\alpha = 5 \pi \frac{r a d}{s} ^ 2$

Now that we have values for $I$ and $\alpha$, we can calculate the torque.

$\tau = I \alpha$

$\tau = \frac{50}{12} k g {m}^{2} \cdot 5 \pi \frac{r a d}{s} ^ 2$

$\tau = \frac{250 \pi}{12} N m$

tau≈65Nm