Question

What torque would have to be applied to a rod with a length of `8 m` and a mass of `8 kg` to change its horizontal spin by a frequency `1 Hz` over `2 s`?

Answer 1

The torque, for the rod rotating about the center is `=134.04Nm`
The torque, for the rod rotating about one end is `=536.17Nm`

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*8*8^2=42.67 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(1)/2*2pi`

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

So the torque is `tau=42.67*(pi) Nm=134.04Nm`

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

`I=1/3*mL^2`

`=1/3*8*8^2=170.67`

So,

The torque is `tau=170.67*(pi)=504/5pi=536.17Nm`