Question

What torque would have to be applied to a rod with a length of `6 m` and a mass of `2 kg` to change its horizontal spin by a frequency `7 Hz` over `8 s`?

Answer 1

The torque for the rod rotating about the center is `=33Nm`
The torque for the rod rotating about one end is `=132Nm`

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=2kg`

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

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

The rate of change of angular velocity is

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

`=(7/4pi) rads^(-2)`

So the torque is `tau=6*(7/4pi) Nm=21/2piNm=33Nm`

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

`I=1/3*mL^2`

`=1/3*2*6^2=24kgm^2`

So,

The torque is `tau=24*(7/4pi)=42pi=132Nm`