Pictured on the right is a bar which has a length of L = 3.48 m. This diagram is from above looking down on the situation. The bar is free to rotate about a frictionless bearing represented by the dot in the center of the bar. Two forces are applied to the two ends as shown.F₁ = 1.86 N and F₂ = 3.18 N. This causes an angular acceleration of a 0.642 rad/2 (CCW). = What is the moment of inertia for the bar? Select one: a. b. OC. 5.04 kg m² 27.3 kg m² 2 13.7 kg m² d. 8.62 kg m² F1 Z|7| L L F₂

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Chapter10: Fixed-axis Rotation
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Pictured on the right is a bar which has a length of
L = 3.48 m . This diagram is from above looking down
on the situation. The bar is free to rotate about a
frictionless bearing represented by the dot in the center
of the bar. Two forces are applied to the two ends as
shown.F₁ = 1.86 N and F₂ = 3.18 N. This causes an
angular acceleration of a = 0.642 rad/2 (CCW).
What is the moment of inertia for the bar?
Select one:
a.
b.
5.04 kg m²
27.3 kg m²
C.
13.7 kg m²
O d. 8.62 kg m²
L
2
F₂
Transcribed Image Text:Pictured on the right is a bar which has a length of L = 3.48 m . This diagram is from above looking down on the situation. The bar is free to rotate about a frictionless bearing represented by the dot in the center of the bar. Two forces are applied to the two ends as shown.F₁ = 1.86 N and F₂ = 3.18 N. This causes an angular acceleration of a = 0.642 rad/2 (CCW). What is the moment of inertia for the bar? Select one: a. b. 5.04 kg m² 27.3 kg m² C. 13.7 kg m² O d. 8.62 kg m² L 2 F₂
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