Question

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

Answer 1

The torque is `=1.05Nm`

Explanation:

The torque is the rate of change of angular momentum

`tau=(dL)/dt=(d(Iomega))/dt=I(domega)/dt`

The is `I=1/12ml^2`

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

The rate of change of angular velocity is

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

`=2pi rads^(-2)`

So the torque is `tau=1/6*2pi Nm=pi/3Nm=1.05Nm`