Question

What torque would have to be applied to a rod with a length of `7 m` and a mass of `9 kg` to change its horizontal spin by a frequency `12 Hz` over `16 s`?

Answer 1

The torque (about the center) is `=173.2Nm`
The torque (about one end) is `=692.7Nm`

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 abour the center is

`I=1/12*mL^2`

`=1/12*9*7^2= 147/4 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(12)/16*2pi`

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

So the torque is `tau=147/4*(3pi)/2 Nm=441/8piNm=173.2Nm`

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

`I=1/3*mL^2`

`=1/3*9*7^2=147`

So,

The torque is `tau=147*(3/2pi)=441/2pi=692.7Nm`