What torque would have to be applied to a rod with a length of 4 m and a mass of 2 kg to change its horizontal spin by a frequency 8 Hz over 8 s?
1 Answer
Feb 9, 2016
- If the rod spins about an axis passing through its center:
\tau = I\frac{d\omega}{dt}=(8/3 kg.m^2)(2\pi (rad)/s)=\frac{16\pi}{3} N.m - If the rod spins about an axis passing through its end:
\tau = I\frac{d\omega}{dt}=(32/3 kg.m^2)(2\pi (rad)/s)=\frac{64\pi}{3} N.m
Explanation:
Torque:
The moment-of-inertia
- If the rod spins about an axis passing through its centre,
I=\frac{1}{12}ML^2 = \frac{1}{12}(2kg)(4m)^2=8/3 kg.m^2
\tau = I\frac{d\omega}{dt}=(8/3 kg.m^2)(2\pi (rad)/s^2)=\frac{16\pi}{3} N.m - If the rod spins about an axis passing through its end,
I=\frac{1}{3}ML^2 = \frac{1}{3}(2kg)(4m)^2=\frac{32}{3} kg.m^2
\tau = I\frac{d\omega}{dt}=(32/3 kg.m^2)(2\pi (rad)/s^2)=\frac{64\pi}{3} N.m