Question

What torque would have to be applied to a rod with a length of `3 m` and a mass of `4 kg` to change its horizontal spin by a frequency `7 Hz` over `1 s`?

Answer 1

The torque (for the rod rotating about the center) is `=131.95Nm`
The torque (for the rod rotating about one end) is `=527.8Nm`

Explanation:

The torque is the rate of change of angular momentum

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

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

`I=1/12*mL^2`

`=1/12*4*3^2= 3 kgm^2`

The rate of change of angular velocity is

`(domega)/dt=(7)/1*2pi`

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

So the torque is `tau=3*(14pi) Nm=42piNm=131.95Nm`

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

`I=1/3*mL^2`

`=1/3*4*3^2=12kgm^2`

So,

The torque is `tau=12*(14pi)=527.8Nm`