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 `7 Hz` over `4 s`?

Answer 1

The torque for the rod rotating about the center is `=469.1Nm`
The torque for the rod rotating about one end is `=1876.6Nm`

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=8kg`

The length of the rod is `L=8m`

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=(7)/4*2pi`

`=(7/2pi) rads^(-2)`

So the torque is `tau=42.67*(7/2pi) Nm=469.1Nm`

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

`I=1/3*mL^2`

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

So,

The torque is `tau=170.67*(7/2pi)=1876.6Nm`