# 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#