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bagirrra123 [75]
3 years ago
10

A person of mass 80 kg stands at the center of a rotating merry-go-round platform of radius 3.5 m and moment of inertia 950 kg*m

^2 . The platform rotates without friction with angular velocity 0.85 rad/s . The person walks radially to the edge of the platform.
1. Calculate the angular velocity when the person reaches the edge.
w=______________ rad/s
2. Calculate the rotational kinetic energy of the system of platform plus person before the person's walk.
Ki=____________ J
3. Calculate the rotational kinetic energy of the system of platform plus person after the person's walk.
Kf=__________ J
Physics
1 answer:
Leya [2.2K]3 years ago
3 0

Answer with Explanation:

We are given that

Mass of person,M=80 kg

Radius,r=3.5 m

Moment of inertia of merry,I=950 kgm^2

Angular velocity of platform=\omega=0.85rad/s

1.Law of conservation of angular momentum

I\omega=I'\omega'

WhereI'=950+80(3.5)^2

Substitute the values

950\times 0.85=(950+80(3.5)^2)\omega'

\omega'=\frac{950\times 0.85}{950+80(3.5)^2}=0.418rad/s

Angular velocity when the person reaches the edge =0.418 rad/s

2.Rotational kinetic energy of the system  before the person's walk

K_i=\frac{1}{2}I\omega^2=\frac{1}{2}(950)(0.85)^2=343.2 J

3.Rotational kinetic energy of the system  after the person's walk

K_f=\frac{1}{2}(950+80(3.5)^2)\times (0.418)^2

K_f=168.6 J

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we substitute in the first expression

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6 0
3 years ago
MathPhys Pls see this Thank you in Advance MathPhys Is the best
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Answer:

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Sum the forces in the perpendicular direction:

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N − mg cos θ = 0

N = mg cos θ

The block is at rest, so F = N μs:

F = N μs

F = mg μs cos θ

F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)

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Sum the forces in the parallel direction (down the incline is positive):

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Sum the forces in the parallel direction (this time, acceleration is not 0).

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4 0
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3 0
4 years ago
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