Question

A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?

Answer 1

The rod's mass moment of inertia is 5kgm².

Moment of Inertia:

The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.

The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.

If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;

or = M / L = dm / dl

dm = (M / L) dl

$I = \int\limits^M_0 {r^2} \, dm$

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.

$I = \int\limits^\frac{L}{2} _\frac{-L}{2} {I^2 (M/L)} \, dl$

$I = \frac{M}3L}[(\frac{L^3}{2^3} - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2$

Mass of the rod = 15 kg

Length of the rod = 2.0 m

Moment of Inertia, I = $\frac{1}{12}15 (2)^2$

                               = 5 kgm²

Therefore, the moment of inertia is 5kgm².

Learn more about moment of inertia here:

brainly.com/question/14119750

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