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#?

1 Answer
Feb 28, 2016

Answer:

#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}#.