Question

What torque would have to be applied to a rod with a length of `6 m` and a mass of `3 kg` to change its horizontal spin by a frequency `15 Hz` over `6 s`?

Answer 1

The torque, for the rod rotating about the center, is `=141.4Nm`
The torque, for the rod rotating about one end, is `=565.5Nm`

Explanation:

The torque is the rate of change of angular momentum

`tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt`

The moment of inertia of a rod, rotating about the center is

`I=1/12*mL^2`

`=1/12*3*6^2= 9 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(15)/6*2pi`

`=(5pi) rads^(-2)`

So the torque is `tau=9*(5pi) Nm=45piNm=141.4Nm`

The moment of inertia of a rod, rotating about one end is

`I=1/3*mL^2`

`=1/3*3*6^2=36`

So,

The torque is `tau=36*(5pi)=180pi=565.5Nm`