Question

What torque would have to be applied to a rod with a length of `1 m` and a mass of `1 kg` to change its horizontal spin by a frequency of `7 Hz` over `3 s`?

Answer 1

The torque for the rod rotating about the center is `=1.22Nm`
The torque for the rod rotating about one end is `=4.89Nm`

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

The rate of change of angular velocity is

`(domega)/dt=(7)/3*2pi`

`=(14/3pi) rads^(-2)`

So the torque is `tau=1/12*(14/3pi) Nm=7/18piNm=1.22Nm`

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

`I=1/3*mL^2`

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

So,

The torque is `tau=1/3*(14/3pi)=14/9pi=4.89Nm`