Conceptual Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 130 kg-m². When one person is standing at a distance of 1.58 m from the center, the carousel has an angular velocity of 0.552 rad/s. However, as this person moves inward to a point located 0.748 m from the center, the angular velocity increases to 0.811 rad/s. What is the person's mass? Number i Units

Principles of Physics: A Calculus-Based Text
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Author:Raymond A. Serway, John W. Jewett
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Chapter10: Rotational Motion
Section: Chapter Questions
Problem 10P: A wheel 2.00 m in diameter lies in a vertical plane and rotates about its central axis with a...
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I'm pretty sure I got the first step right, but I'm really struggling with the second step. I think I set up the equation correctly, but when it comes to actually solving, I'm not getting the steps right. Can you please list out step by step how to solve the equation in part 2, assuming I got that equation right. I've attached the problem, as well as my work.

Conceptual Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on
frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 130 kg-m². When
one person is standing at a distance of 1.58 m from the center, the carousel has an angular velocity of 0.552 rad/s. However, as this
person moves inward to a point located 0.748 m from the center, the angular velocity increases to 0.811 rad/s. What is the person's
mass?
Number i
eTextbook and Media
Units
Transcribed Image Text:Conceptual Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 130 kg-m². When one person is standing at a distance of 1.58 m from the center, the carousel has an angular velocity of 0.552 rad/s. However, as this person moves inward to a point located 0.748 m from the center, the angular velocity increases to 0.811 rad/s. What is the person's mass? Number i eTextbook and Media Units
A
at
playground carousel is free to rotate about it's center on frictionless
bearings, and air resistence is
has a moment of inertia of 130 kg.m² When one
negligable. The carousel itself (without riders)
a distance of 1.58 m from the center, the carousel has an angular velocity
person is standing.
of.552 rad/s. However, as this person
mores inward to a point =
located . 148m from the center, the angular velocity
What is the
eson
227
the angolar velocity recreased to - Ellrad/s
persons mass?
I: 130kg.m²
Xo:1.58m
w..552 rad /s
x: 748m
Wp: 811 rad /s
find H
CQUATIONS
Conservation of Angular Homendum: I₁ W₁=I₂ W₂
Parallel Axis I "Ie +mr²
ⒸI₁ = 130 kg⋅ m² + m (1.58)²
I 13
+ 2.4964m
130 куне
I ₂ = 130 kg. m² + m( . 748m) ²
-
mida to
130kg.m² + 1.378m²= 130 + .45375m
131.375m
130.45375m
de 2004
dan A
ERS
I ₂ = 130 kg.m² +₁5595m - poh 45
T₂W₁ = I₂ W₂
LA
(130 kg⋅m² + 2.4964 m) (.552) = (130kg⋅m² +.5595m) (811)
SOLVE FOR H
20
35
E
Transcribed Image Text:A at playground carousel is free to rotate about it's center on frictionless bearings, and air resistence is has a moment of inertia of 130 kg.m² When one negligable. The carousel itself (without riders) a distance of 1.58 m from the center, the carousel has an angular velocity person is standing. of.552 rad/s. However, as this person mores inward to a point = located . 148m from the center, the angular velocity What is the eson 227 the angolar velocity recreased to - Ellrad/s persons mass? I: 130kg.m² Xo:1.58m w..552 rad /s x: 748m Wp: 811 rad /s find H CQUATIONS Conservation of Angular Homendum: I₁ W₁=I₂ W₂ Parallel Axis I "Ie +mr² ⒸI₁ = 130 kg⋅ m² + m (1.58)² I 13 + 2.4964m 130 куне I ₂ = 130 kg. m² + m( . 748m) ² - mida to 130kg.m² + 1.378m²= 130 + .45375m 131.375m 130.45375m de 2004 dan A ERS I ₂ = 130 kg.m² +₁5595m - poh 45 T₂W₁ = I₂ W₂ LA (130 kg⋅m² + 2.4964 m) (.552) = (130kg⋅m² +.5595m) (811) SOLVE FOR H 20 35 E
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