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
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^2I=112MR2
= 1/12 (2 "kg") (1 "m")^2=112(2kg)(1m)2
= 0.17 "kgm"^{-2}=0.17kgm−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