Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `3 kg` to change its horizontal spin by a frequency of `5 Hz` over `2 s`?

Answer 1

The torque for the rod rotating about its center is `=62.83Nm`
The torque for the rod rotating about one end is `=251.33Nm`

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=3kg`

The length of the rod is `L=4m`

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

`I=1/12*mL^2`

`=1/12*3*4^2= 4 kgm^2`

The frequency is `f=5 Hz`

The rate of change of angular velocity is

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

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

So the torque is `tau=4*(5pi) Nm=20piNm=62.83Nm`

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

`I=1/3*mL^2`

`=1/3*3*4^2=16kgm^2`

So,

The torque is `tau=16*(5pi)=80pi=251.33Nm`