Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `3 kg` to change its horizontal spin by a frequency of `12 Hz` over `4 s`?

Answer 1

The torque, for the rod rotating about the center, is `=75.4Nm`
The torque, for the rod rotating about one end, is `=30.1.6Nm`

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*3*4^2= 4 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(12)/4*2pi`

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

So the torque is `tau=4*(6pi) Nm=24piNm=75.4Nm`

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

`I=1/3*mL^2`

`=1/3*3*4^2=16`

So,

The torque is `tau=16*(6pi)=96pi=301.6Nm`