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vladimir1956 [14]
3 years ago
12

Max (15 kg) and maya (12 kg) are riding on a merry-go-round that rotates at a constant speed. max is sitting on the edge of the

merry-go-round, 2.4 m from the center, and maya is 1.2 m from the center. considering max and maya to be one system of masses, what is their moment of inertia measured with respect to the center of the merry-go-round?
Physics
2 answers:
insens350 [35]3 years ago
8 0

Answer:

103.68 NM

Explanation:

<u>Step 1:</u> identify the given parameters

Max mass = 15kg

maya mass = 12kg

max distance from the center (a)= 2.4m

maya distance from the center (b) = 1.2m

Moment of inertia = M*r²

<u>Step 2:</u> calculate the moment of inertia (I) of max

I = M*r²

I = 15*(2.4)² = 86.4NM

<u>Step 3:</u> calculate moment of inertia of Maya

I = M*r²

I = 12*(1.2)² = 17.28NM

<u>Step 4: </u>find the total moment of inertia

Total moment of inertia = max moment of inertia + Maya moment of inertia

Total moment of inertia = 86.4 NM + 17.28NM

Total moment of inertia = 103.68 NM

Lana71 [14]3 years ago
5 0
1.2 GLAD TO HELP MY FRIEND
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The magnitude of the forces acting at the top are;

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The location of the point the person is standing = 3 meters from the bottom

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The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

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Taking moment about the point of contact of the ladder with the ground, <em>B </em>gives;

\sum M_B = 0

Therefore;

\sum M_{BCW} = \sum M_{BCCW}

Where;

\sum M_{BCW} = The sum of clockwise moments about <em>B</em>

\sum M_{BCCW} = The sum of counterclockwise moments about <em>B</em>

Therefore, we have;

\sum M_{BCW} = 2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

\sum M_{BCCW} = F_R × √(6² - 2²)

Therefore, we get;

2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81  = F_R × √(6² - 2²)

F_R  = (2  × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

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We note that the magnitude of the reaction force at the roof, F_R = The magnitude of the frictional force of bottom of the ladder on the floor, F_f but opposite in direction

Therefore;

F_R = -F_f

F_f = - F_R ≈ -132.95 N

Similarly, at equilibrium, we have;

∑Fₓ = \sum F_y = 0

The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;

\sum F_y = -70.0 × 9.81 - 10 × 9.81 + F_{By}

∴ The upward force acting at the bottom, F_{By} = 784.8 N

Therefore;

The magnitudes of the forces at the ladder top and bottom are;

At the top;

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\mathbf{F_{Bottom, \ x}} = F_f ≈ -132.95 N →

\mathbf{F_{Bottom, \ y}} = F_{By} = 784.8 N ↑

Learn more about equilibrium of forces here;

brainly.com/question/16051313

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