Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `2 kg` to change its horizontal spin by a frequency `8 Hz` over `8 s`?

Answer 1

1. If the rod spins about an axis passing through its center:
   `\tau = I\frac{d\omega}{dt}=(8/3 kg.m^2)(2\pi (rad)/s)=\frac{16\pi}{3} N.m`
2. If the rod spins about an axis passing through its end:`\tau = I\frac{d\omega}{dt}=(32/3 kg.m^2)(2\pi (rad)/s)=\frac{64\pi}{3} N.m`

Explanation:

Torque: `\tau \equiv \frac{dL}{dt}= I\alpha = I\frac{d\omega}{dt}`

`L` - Angular Momentum; `\quad` `I` - Moment-of-inertia;
`alpha` - Angular acceleration; `\quad` `\omega` - Angular frequency;
`f` - frequency.

`\omega = 2\pi f; \qquad \frac{d\omega}{dt} = 2\pi\frac{df}{dt}=2\pi\frac{8Hz}{8s}=2\pi` `(rad)/s^2`.

The moment-of-inertia `I` of the rod depends on the location of the spin axis. If `M` is the mass of the rod and `L` its length then the moment-of-inertia of the rod is:

`M = 2` kg `\qquad` `L = 4` m

1. If the rod spins about an axis passing through its centre,
   `I=\frac{1}{12}ML^2 = \frac{1}{12}(2kg)(4m)^2=8/3 kg.m^2`
   `\tau = I\frac{d\omega}{dt}=(8/3 kg.m^2)(2\pi (rad)/s^2)=\frac{16\pi}{3} N.m`
2. If the rod spins about an axis passing through its end,
   `I=\frac{1}{3}ML^2 = \frac{1}{3}(2kg)(4m)^2=\frac{32}{3} kg.m^2`
   `\tau = I\frac{d\omega}{dt}=(32/3 kg.m^2)(2\pi (rad)/s^2)=\frac{64\pi}{3} N.m`