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Lilit [14]
3 years ago
15

A 5.0-m radius playground merry-go-round with a moment of inertia of 2000 kg · m 2 is rotating freely with an angular speed of 1

.0 rad/s. Two people, each having a mass of 60 kg, are standing right outside the edge of the merry-go-round and step on it with negligible speed. What is the angular speed of the merry-go-round right after the two people have stepped on?
Physics
1 answer:
lidiya [134]3 years ago
5 0

Answer:

angular speed = 0.4 rad/s

Explanation:

given data

radius = 5 m

moment of inertia = 2000 kg-m²

angular speed = 1.0 rad/s

mass = 60 kg

to find out

angular speed

solution

Rotational momentum of merry-go-round = I?

we get here momentum that is express as  

momentum = 2000 × 1

momentum = 2000 kg-m²/s

and  

Inertia of people will be here as

Inertia of people = mr² = 60 × 5²

Inertia of people = 1500 kg-m²

so Inertia of people for two people  

1500 × 2 = 3000

and

now conserving angular momentum(ω)

moment of inertia × angular speed = ( momentum + Inertia of people ) angular momentum  

2000 ×  1 = (2000 + 3000 ) ω

solve we get now  

ω = 0.4 rad/s  

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Answer:

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\frac{Q}{A} =\frac{T_1-T_0}{d_w}k_w=\frac{T_2-T_1}{d_s}k_s\\T_1(\frac{k_w}{d_w}+\frac{k_s}{d_s})=T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}\\T_1=\frac{T_2\frac{k_s}{d_s}+T_0\frac{k_w}{d_w}}{\frac{k_w}{d_w}+\frac{k_s}{d_s}}\\T_1= 15.54 \°C

Then, to find the rate of heat flow per square meter, we have:

\frac{Q}{A}=\frac{T_1-T_0}{d_w}k_w=0.119 \frac{W}{m^2}\\\frac{Q}{A}=\frac{T_2-T_1}{d_s}k_s= 0.119 \frac{W}{m^2}

T_0: Temperature \ in \ the \ house\\T_1: Temperature \ at \ the \ plane \ between \ wood \ and \ styrofoam\\T_2: Temperature \ outside\\k_w: k \ for \ wood\\d_w: wood \ thickness\\k_s: k \ for \ styrofoam\\d_s: styrofoam \ thickness

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