Respuesta :
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
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
- Impacts of Gravity : https://brainly.com/question/5330244
- 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
