Question

What torque would have to be applied to a rod with a length of `5 m` and a mass of `12 kg` to change its horizontal spin by a frequency `8 Hz` over `7 s`?

Answer 1

The torque for the rod rotating about the center is `=179.5Nm`
The torque for the rod rotating about one end is `=718.1Nm`

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`

The mass of the rod is `m=12kg`

The length of the rod is `L=5m`

`=1/12*12*5^2= 25 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(8)/7*2pi`

`=(16/7pi) rads^(-2)`

So the torque is `tau=25*(16/7pi) Nm=400/7piNm=179.5Nm`

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

`I=1/3*mL^2`

`=1/3*12*5^2=100kgm^2`

So,

The torque is `tau=100*(16/7pi)=1600/7pi=718.1Nm`