Question

What torque would have to be applied to a rod with a length of `1 m` and a mass of `2 kg` to change its horizontal spin by a frequency of `7 Hz` over `2 s`?

Answer 1

`tau_"ave" = 0.583 "kgm"^{-2}"s"^{-2}`

Explanation:

I'm assuming that the rod is spinning about its center.

The moment of inertia of the rod about its center, `I`, is given by

`I = 1/12 M R^2`

`= 1/12 (2 "kg") (1 "m")^2`

`= 0.17 "kgm"^{-2}`

The change in angular momentum is given by

`I Delta omega = (0.17 "kgm"^{-2})(7 "Hz")`

`= 1.17 "kgm"^{-2}"s"^{-1}`

The angular impulse is given by

`tau_"ave" * t = tau_"ave" * (2 "s")`

According to the Momentum-Impulse Theorem, the change in angular momentum is equal to the angular impulse. Therefore, we can write

`1.17 "kgm"^{-2}"s"^{-1} = tau_"ave" * (2 "s")`

`tau_"ave" = 0.583 "kgm"^{-2}"s"^{-2}`

Note that the instantaneous torque does not always have to be `0.583 "kgm"^{-2}"s"^{-2}`. It is just that the time-averaged torque that has to be `0.583 "kgm"^{-2}"s"^{-2}`.