Question

What torque would have to be applied to a rod with a length of `6 m` and a mass of `8 kg` to change its horizontal spin by a frequency `8 Hz` over `5 s`?

Answer 1

The torque for the rod rotating about the center is `=241.3Nm`
The torque for the rod rotating about one end is `=965.1Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The mass of the rod is `m=8kg`

The length of the rod is `L=6m`

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

`I=1/12*mL^2`

`=1/12*8*6^2= 24 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(8)/5*2pi`

`alpha=(16/5pi) rads^(-2)`

So the torque is `tau=24*(16/5pi) Nm=241.3Nm`

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

`I=1/3*mL^2`

`=1/3*8*6^2=96kgm^2`

So,

The torque is `tau=96*(16/5pi)=504/5pi=965.1Nm`