What is the moment of inertia of a pendulum with a mass of 5 kg that is 3 m from the pivot?

1 Answer
MetaPhysik
Dec 24, 2017

45 kgm^2

Explanation:

Just use the definition of moment of inertia
I = mr^2 = 5 kg* (3m)^2 = 45 kgm^2

To appreciate this definition, take a look at its kinetic energy at any given constant,

K = 1/(2)mv^2

Since a pendulum swings back and forth, you can measure/observe two kinds of speeds, the speed of the blob (v) and the speed of the angular change over time (omega). Their relationship is

v=romega

Then K = 1/(2) m (romega)^2 =1/(2) (m r^2) omega^2

Let I = 1/(2) mr^2 and call it moment of inertia

K =1/(2) I omega^2

From an utility point of view, it is much easier to measure the angular change over time (omega) then the actual speed (v) of the blob. The trade off is that besides knowing the mass, you also need to know the radius of rotation (length of the pendulum) I in order to figure out the kinetic energy of the pendulum.

And of course, you can use the concept of angular momentum to appreciate the definition of I as well.

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