Question

What torque would have to be applied to a rod with a length of `6 m` and a mass of `3 kg` to change its horizontal spin by a frequency `5 Hz` over `4 s`?

Answer 1

`tau=657.75` `N-m`

Explanation:

The torque is given by:
`tau=I*dot omega=I*omega/t`
where `I` is the moment of Inertia of the horizontal spinning rod which is given by: `I=(m*l^2)/12`
and `omega`, the angular velocity is given by `omega=2pi*f`

So, from the given data
`I=(3*6^2)/12=9` `kg-m^2`
`omega=2pi * 5=31.4` `(rad)/s`
`t=4` `s`

Thus we can now evaluate the torque
`tau=I*omega/t`
`tau=9*31.4/4=657.75` `N-m`