Question

What torque would have to be applied to a rod with a length of `3 m` and a mass of `1 kg` to change its horizontal spin by a frequency of `5 Hz` over `2 s`?

Answer 1

The torque for the rod rotating about the center is `=11.78Nm`
The torque for the rod rotating one end is `=47.12Nm`

Explanation:

The torque is the rate of change of angular momentum

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

The mass of the rod is `m=1kg`

The length of the rod is `L=3m`

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

`I=1/12*mL^2`

`=1/12*1*3^2= 0.75 kgm^2`

The rate of change of angular velocity is

`alpha=(domega)/dt=(5)/2*2pi`

`=(5pi) rads^(-2)`

So the torque is `tau=0.75*(5pi) Nm=11.78Nm`

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

`I=1/3*mL^2`

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

So,

The torque is

`=3*(5pi)=47.12Nm`