Question

What torque would have to be applied to a rod with a length of `7 m` and a mass of `12 kg` to change its horizontal spin by a frequency `15 Hz` over `9 s`?

Answer 1

The torque for the rod rotating about the center is `=513.1Nm`
The torque for the rod rotating about one end is `=2052.5Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The mass of the rod is `m=12kg`

The length of the rod is `L=7m`

The moment of inertia of a rod, rotating about the center is

`I=1/12*mL^2`

`=1/12*12*7^2= 49 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(15)/9*2pi`

`=(10/3pi) rads^(-2)`

So the torque is `tau=49*(10/3pi) Nm=513.1Nm`

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

`I=1/3*mL^2`

`=1/3*12*7^2=196kgm^2`

So,

The torque is `tau=196*(10/3pi)=2052.5Nm`