What torque would have to be applied to a rod with a length of 1 m1m and a mass of 2 kg2kg to change its horizontal spin by a frequency of 7 Hz7Hz over 2 s2s?

1 Answer
Feb 28, 2016

tau_"ave" = 0.583 "kgm"^{-2}"s"^{-2}τave=0.583kgm2s2

Explanation:

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

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

I = 1/12 M R^2I=112MR2

= 1/12 (2 "kg") (1 "m")^2=112(2kg)(1m)2

= 0.17 "kgm"^{-2}=0.17kgm2

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}.