Question

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`?

Answer 1

The angular acceleration is `(3pi) text{radians} * 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 = text{change in angular velocity} / text{time}`

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 * `text{rod length}^2`
so:
moment of inertia = 2kg * `1m^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