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

The mass of the rod is 16 kg.

Step-by-step 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=(ML^2)/(12)`

Where

M is mass of the rod

`M=(12I)/(L^2)\\\\M=(12* 12)/((3)^2)\\\\M=16\ kg`

So, the mass of the rod is 16 kg.