Question

Consider a long, thin rod with a length of 3 m rotating about it's end. This rod has a moment of inertia of 12 kg·m2 about this pivot.
What is the mass of the rod? Give your answer in kilograms to two decimal places.

Answer 1

Answer:

The mass of the rod is 16 kg.

Explanation:

Given that,

The length of a rod, L = 3 m

The moment of inertia of the rod, I = 12 kg-m²

We need to find the mass of the rod. The moment of inertia of the rod of length L is given by :

$I=\dfrac{ML^2}{12}$

Where

M is mass of the rod

$M=\dfrac{12I}{L^2}\\\\M=\dfrac{12\times 12}{(3)^2}\\\\M=16\ kg$

So, the mass of the rod is 16 kg.