Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `3 kg` to change its horizontal spin by a frequency of `12 Hz` over `1 s`?

Answer 1

Torque (`tau`) is defined as change in rate of angular momentum (`L`)i.e `(dL)/dt` i.e `(d(Iomega))/dt` i.e `I (domega)/dt` (where,`I`nis the moment of inertia)

Now,moment of inertia of a rod w.r.t its one end along an axis perpendicular to its plane is `(ML^2)/3` i.e `3*(4)^2/3` or `16 Kg.m^2`

Now, rate of change in angular velocity = `(domega)/dt` = `2pi (nu1-nu2)/t` i.e `2 pi*12/1` or `24 pi`

So, `tau = 16* 24 pi` or, `1206.37 N.m`