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ANEK [815]
3 years ago
13

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.
Physics
1 answer:
Orlov [11]3 years ago
5 0

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.

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