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jolli1 [7]
3 years ago
15

A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

s that lies in the plane of the plate, passes through the center of the plate, and is parallel to the side with length b.
Physics
1 answer:
slega [8]3 years ago
4 0
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



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Answer:

83.6°

Explanation:

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So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

1.5 × sinθ₁  = 1

sinθ₁ = 1/1.5

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So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.

5 0
3 years ago
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The answer is 18000 J

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5 0
2 years ago
weight of Ali is 500andN.he is standing on the ground with an area of 0.025 m^2 area .we can find pressure under his feet. what
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Answer:

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8 0
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3 years ago
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