# 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?

Feb 28, 2016

${\tau}_{\text{ave" = 0.583 "kgm"^{-2}"s}}^{- 2}$

#### 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

$I = \frac{1}{12} M {R}^{2}$

$= \frac{1}{12} {\left(2 \text{kg") (1 "m}\right)}^{2}$

$= 0.17 {\text{kgm}}^{- 2}$

The change in angular momentum is given by

$I \Delta \omega = \left(0.17 \text{kgm"^{-2})(7 "Hz}\right)$

$= 1.17 {\text{kgm"^{-2}"s}}^{- 1}$

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")

${\tau}_{\text{ave" = 0.583 "kgm"^{-2}"s}}^{- 2}$

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