Question

What torque would have to be applied to a rod with a length of `4 m` and a mass of `3 kg` to change its horizontal spin by a frequency of `3 Hz` over `2 s`?

Answer 1

The torque applied to the rod is 37.68 Newton-metre.

Explanation:

Torque is defined as the product of moment of inertia and angular acceleration
`=>tau=Ialpha`
Moment of inertia of rod is `I_"rod"=((ML^2)/12)` and angular acceleration is change in angular velocity with respect to time
`=>alpha=omega/t`
angular velocity `omega=2pinu` as frequency is given in above case.
Now the torque becomes `tau=[(ML^2)/12][(2pinu)/t]`
Substituting the given values we get,
`tau=[[(3kg)(4m)^2]/12][(2xx3.14xx3Hz)/(2s)]`
`:.`torque`tau=37.68`N-m.