Question

Find the moment of inertia of a rectangle of uniform mass m with a width h and length l about z axis

Answer 1

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Imagining the rod is cut into infinitesimally many pieces of infinitesimally thin slices. (“slicing” the object)

Each of these slices have a mass of dm (remember your calculus lessons?) and length of dx.

Choose a variable to sum. E.g. In this problem, we are summing from left of the axis to right of axis. The variable is x. In other problem, the variable can be theta or r or …

Now, we show our formula for the calculation for moment of inertia first:

dI=dmx2dI=dmx2

Hey, there is a dm in the equation! Recall that we’re using x to sum. Hence, we have to force a dx into the equation for moment of inertia. Now, lets find an expression for dm. Since the rod is uniform, the mass varies linearly with distance.

dm=MLdxdm=MLdx

Using the equation for dm, we substitute it into the first equation. Hence, we have:

dI=MLx2dxdI=MLx2dx

Now,

I=∫dII=∫dI

Substituting dI, (write the appropriate limits)

I=ML∫−hL–hx2dxI=ML∫−hL–hx2dx

Note: The lower limit is -h because the left side of the rod is -h units away from the axis of rotation. (We take right as positive)

Solving the integration, we have:

I=13M(L2–3Lh+3h2)I=13M(L2–3Lh+3h2)

We’re done!

Now, for the special cases,



When the rotation axis is at one end of the rod (h = 0), we have:

I=13ML2I=13ML2

However, if the rotation axis is through the centre of mass of the rod. (remember rod is uniform, hence h=L2h=L2)



We have:

I=112ML2I=112ML2

Note: The moment of inertia is expected to be highest when the axis is at one end since the mass are now furthest away from the axis of rotation. Lowest is when axis is at the center.

– Always check your expression after deriving them.