Question

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

Answer 1

The torque, for the rod rotating about the center, is `=0.35Nm`
The torque, for the rod rotating about one end, is `=1.40Nm`

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

The rate of change of angular velocity is

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

`=(4/3pi) rads^(-2)`

So the torque is `tau=1/12*(4/3pi) Nm=1/9piNm=0.35Nm`

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

`I=1/3*mL^2`

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

So,

The torque is `tau=1/3*(4/3pi)=4/9pi=1.40Nm`