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

Answer:

  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#