Question

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 `9 Hz` over `3 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 `2*(5)^2/3` or `50/3 Kg.m^2`

Now, as the radius of the circle isn't mentioned,so the rod can move about its mid point as well,in that case `I = (ML^2)/12` i.e `25/6 Kg.m^2`

Now, rate of change in angular velocity = `(domega)/dt` = `2pi (nu1-nu2)/t` i.e `2 pi*9/3` or `6 pi`

So, `tau = 50/3* 6 pi` or, `314.16 N.m` or, `25/6*6pi` i.e `78.54 N.m` for the `2nd` case