Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `5 kg` to change its horizontal spin by a frequency of `3 Hz` over `7 s`?

Answer 1

The torque for the rod rotating about the center is `=17.95Nm`
The torque for the rod rotating about one end is `=71.81Nm`

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`

`=1/12*5*4^2= 20/3 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(3)/7*2pi`

`=(6/7pi) rads^(-2)`

So the torque is `tau=20/3*(6/7pi) Nm=40/7piNm=17.95Nm`

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

`I=1/3*mL^2`

`=1/3*5*4^2=80/3kgm^2`

So,

The torque is `tau=80/3*(6/7pi)=160/7pi=71.81Nm`