What is the moment of inertia of a pendulum with a mass of #8 kg# that is #1 m# from the pivot?

Question

1467 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-23T14:32:58

#I=8color(white)(l)"kg"*"m"^2#

Taking the pendulum bulb as a point mass and the pivot as the rotational axis,

#m_"bulb"=8color(white)(l)"kg"#
#r=1color(white)(l)"m"#

Moment of inertia of the bulb as a point mass: [1]

#I=m*r^2=8color(white)(l)"kg"*"m"^2#

Note that the momentum of inertia expression can be derived using the parallel axis theorem: [2]

#I=I_"cm"+m*d^2#

Where #I_"cm"=0# for a point mass and #d=r# the axis of rotation distance from the axis passing through the center of mass.

Hence #I=m*d^2=m*r^2#
Reference
[1] List of moments of inertia, the English Wikipedia, https://en.wikipedia.org/wiki/List_of_moments_of_inertia
[2] Parallel axis theorem, the English Wikipedia, https://en.wikipedia.org/wiki/Parallel_axis_theorem