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 `4 s`?

Answer 1

The torque (rod rotating about the center) is `=0.785Nm`
The torque (rod rotating about one end) is `=3.14Nm`

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

The rate of change of angular velocity is

`(domega)/dt=(3)/4*2pi`

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

So the torque is `tau=1/6*(3/2pi) Nm=1/4piNm=0.785Nm`

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

`I=1/3*mL^2`

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

So,

The torque is `tau=2/3*(3/2pi)=pi=3.14Nm`