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
  1. 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
  2. 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: \tau \equiv \frac{dL}{dt}= I\alpha = I\frac{d\omega}{dt}

L - Angular Momentum; \quad I - Moment-of-inertia;
alpha - Angular acceleration; \quad \omega - Angular frequency;
f - frequency.

\omega = 2\pi f; \qquad \frac{d\omega}{dt} = 2\pi\frac{df}{dt}=2\pi\frac{8Hz}{8s}=2\pi (rad)/s^2.

The moment-of-inertia I of the rod depends on the location of the spin axis. If M is the mass of the rod and L its length then the moment-of-inertia of the rod is:

M = 2 kg \qquad L = 4 m

  1. 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
  2. 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