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Artemon [7]
3 years ago
7

How do you do this question?

Physics
1 answer:
Arisa [49]3 years ago
5 0

Explanation:

The moment of inertia of a solid sphere about any axis passing through its center is I = 2/5 MR².

The moment of inertia of the solid cylinder about Axis A is I = 1/12 m (3r² + L²).  The moment of inertia about Axis C is I = 1/2 mr².

According to parallel axis theorem, the moment of inertia of an object about an offset axis is I = I₀ + md², where d is the offset.

(a) The moment of inertia of the rod is I = 1/12 m (3r² + L²).

Using parallel axis theorem, the moment of inertia of each sphere is I = 2/5 MR² + M (R + L/2)².

The total moment of inertia is therefore:

I = 1/12 m (3r² + L²) + 2 [2/5 MR² + M (R + L/2)²]

I = 1/12 m (3r² + L²) + 4/5 MR² + 2M (R + L/2)²

(b) The moment of inertia of the right sphere is I = 2/5 MR².

The moment of inertia of the rod is I = 1/12 m (3r² + L²) + m (R + L/2)².

The moment of inertia of the left sphere is I = 2/5 MR² + M (2R + L)².

The total moment of inertia is therefore:

I = 2/5 MR² + 1/12 m (3r² + L²) + m (R + L/2)² + 2/5 MR² + M (2R + L)²

I = 4/5 MR² + 1/12 m (3r² + L²) + m (R + L/2)² + M (2R + L)²

(c) The moment of inertia of each sphere is I = 2/5 MR².

The moment of inertia of the rod is I = 1/2 mr².

The total moment of inertia is:

I = 2/5 MR² + 2/5 MR² + 1/2 mr²

I = 4/5 MR² + 1/2 mr²

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