Max is pushing on the edge (2.4 m from the center) of the merry-go-round to make it rotate. maya is riding on the merry-go-round, sitting 1.2 m from the center. if max increases the merry-go-round angular speed from 0.52 rev/s to 1.5 rev/s in 2.2 s, what is the merry-go-round angular acceleration

Respuesta :

W0lf93
Firstly, the angular acceleration is independent of where Maya is sitting, or if she's sitting on the merry-go-round at all. Max only has to supply more energy. 
Initial angular velocity a1: 0.52 rev/s 
Final angular velocity a2: 1.5 rev/s  
Tim taken t: 2.2s  
Angular acceleration: alpha  
Using kinematic equations for angular motion, (a2) = (a1) + alpha*t. 
This gives alpha as 0.4454 rev/s^2

The merry-go-round angular acceleration is about 2.8 rad/s²

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Further explanation

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

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Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

initial angular speed = ωo = 0.52 rev/s = 1.04π rad/s

final angular speed = ω = 1.5 rev/s = 3π rad/s

time taken = t = 2.2 s

Asked:

angular acceleration = α = ?

Solution:

We will use formula for angular acceleration to solve this problem:

[tex]\alpha = ( \omega - \omega_o ) \div t[/tex]

[tex]\alpha = ( 3 \pi - 1.04 \pi ) \div 2.2[/tex]

[tex]\alpha = 1.96 \pi \div 2.2[/tex]

[tex]\alpha \approx 0.89 \pi \texttt { rad/s}^2[/tex]

[tex]\alpha \approx 2.8 \texttt{ rad/s}^2[/tex]

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Learn more

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  • Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
  • The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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