2) The uniform solid block in the figure has mass 0.172 kg and edge lengths a = 3.5 cm, b = 8.4 cm, and c = 1.4 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces. [4.7 x 104 kg m²] Rotation axis

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Chapter13: Rotation Ii: A Conservation Approach
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Can you please solve question number 2 explaining step by step, on a picture please.

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Rotational Inertia Problems
1) In the diagram below, two particles, each with mass M = 0.85 kg, are fastened to each other, and to a rotation axis at
O, by two thin rods, each with length, d = 5.6 cm and mass M = 1.2 kg. The combination rotates around the rotation
axis with angular speed w = 0.30 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b)
kinetic energy? [(a) 0.023 kg m²; (b) 0.001 J]
2=0.85 kg
= = 0.112m²
m₁ = 0.85
= 0.056
mz
Itotal=IRod +Im, + I m₂ =
m
d
M
M
()
Rod - M Rod= 2(1-2 kg) - 2.4
L = 2(0.056)=112
= // ML ²
히
I total= Modkrodt me,² + m₂ {²
= 0.058 + (0.85) (0.056)²+(0.85) (0.112) ²
= =— (2+4) (₁ 112)²
Stola 0.023 Kg m²
= 0.0101
2) The uniform solid block in the figure has mass 0.172 kg and edge lengths a = 3.5 cm, b = 8.4 cm, and c = 1.4 cm.
Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.
[4.7 x 104 kg m²]
Rotation
axis
3) Four point masses of 1.5 kg each are arranged in a square on massless rods. The length of a side of the square is L
= 0.30 m. What is the rotational inertia about the axis shown? [0.135 kg m²]
L
00
m
M
-Rotation axis
0
Irod about =
end
Transcribed Image Text:Rotational Inertia Problems 1) In the diagram below, two particles, each with mass M = 0.85 kg, are fastened to each other, and to a rotation axis at O, by two thin rods, each with length, d = 5.6 cm and mass M = 1.2 kg. The combination rotates around the rotation axis with angular speed w = 0.30 rad/s. Measured about O, what are the combination's (a) rotational inertia and (b) kinetic energy? [(a) 0.023 kg m²; (b) 0.001 J] 2=0.85 kg = = 0.112m² m₁ = 0.85 = 0.056 mz Itotal=IRod +Im, + I m₂ = m d M M () Rod - M Rod= 2(1-2 kg) - 2.4 L = 2(0.056)=112 = // ML ² 히 I total= Modkrodt me,² + m₂ {² = 0.058 + (0.85) (0.056)²+(0.85) (0.112) ² = =— (2+4) (₁ 112)² Stola 0.023 Kg m² = 0.0101 2) The uniform solid block in the figure has mass 0.172 kg and edge lengths a = 3.5 cm, b = 8.4 cm, and c = 1.4 cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces. [4.7 x 104 kg m²] Rotation axis 3) Four point masses of 1.5 kg each are arranged in a square on massless rods. The length of a side of the square is L = 0.30 m. What is the rotational inertia about the axis shown? [0.135 kg m²] L 00 m M -Rotation axis 0 Irod about = end
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