Question

What torque would have to be applied to a rod with a length of `7 m` and a mass of `9 kg` to change its horizontal spin by a frequency `12 Hz` over `9 s`?

Answer 1

The torque for the rod rotating about the center is `=307.9Nm`
The torque for the rod rotating about one end is `=1231.5Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The mass of the rod is `m=9kg`

The length of the rod is `L=7m`

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

`I=1/12*mL^2`

`=1/12*9*7^2= 36.75 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(12)/9*2pi`

`=(8/3pi) rads^(-2)`

So the torque is `tau=36.75*(8/3pi) Nm=307.9Nm`

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

`I=1/3*mL^2`

`=1/3*9*7^2=147kgm^2`

So,

The torque is `tau=147*(8/3pi)=1231.5Nm`