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
Explanation:
I'm assuming that the rod is spinning about its center.
The moment of inertia of the rod about its center,
#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