# 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 5 Hz over 4 s?

May 20, 2017

tau≈33Nm

#### 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,

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

$= \frac{25}{6} k g {m}^{2}$

The angular acceleration of the rod as it rotates, $\alpha$, can be calculated from the given values of time and frequency, given by:

alpha=(Δomega)/( Δt).

We can find $\omega$ from the given change in frequency of $5 H z$, as $\omega = 2 \pi f$.

$\omega = 2 \pi \left(5 {s}^{-} 1\right)$

$= 10 \pi \frac{r a d}{s}$

Because the problem states that the frequency changed by this amount, this is  Δomega. We are given that this took place over $4 s$, which is our  Δt value. Thus,

$\alpha = \frac{10 \pi \frac{r a d}{s}}{4 s}$

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

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

$\tau = I \alpha$

$= \frac{25}{6} k g {m}^{2} \cdot \frac{5 \pi}{2} \frac{r a d}{s} ^ 2$

$= \frac{125 \pi}{12} N m$

≈33Nm