Question

What torque would have to be applied to a rod with a length of `3 m` and a mass of `1 kg` to change its horizontal spin by a frequency of `2 Hz` over `2 s`?

Answer 1

The torque, for the rod rotating about the center is `=4.71Nm`
The torque, for the rod rotating about one end is `=18.85Nm`

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*1*3^2=0.75 kgm^2`

The rate of change of angular velocity is

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

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

So the torque is `tau=0.75*(2pi) Nm=4.71Nm`

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

`I=1/3*mL^2`

`=1/3*1*3^2=3`

So,

The torque is `tau=3*(2pi)=6pi=18.85Nm`