Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `2 kg` to change its horizontal spin by a frequency `9 Hz` over `2 s`?

Answer 1

The torque for the rod rotating about the center is `=75.49Nm`
The torque for the rod rotating about one end is `=301.69Nm`

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=2kg`

The length of the rod is `L=4m`

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

`I=1/12*mL^2`

`=1/12*2*4^2= 2.67 kgm^2`

The rate of change of angular velocity is

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

`=(9pi) rads^(-2)`

So the torque is `tau=2.67*(9pi) Nm=75.49Nm`

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

`I=1/3*mL^2`

`=1/3*2*4^2=10.67kgm^2`

So,

The torque is `tau=10.67*(9pi)=301.69Nm`