The acceleration of that point due to the spin of the wheel is about 2.0 × 10² m/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:
radius of wheel = R = 40 cm = 0.4 m
angular velocity = ω = 5.0 rev/s = 10π rad/s
distance of the point from axis of rotation = r = 0.20 m
Asked:
acceleration of that point = a = ?
Solution:
[tex]a = \omega^2 r[/tex]
[tex]a = (10 \pi )^2 \times 0.20[/tex]
[tex]a = 20 \pi^2 \texttt{ m/s}^2[/tex]
[tex]a \approx 2.0 \times 10^2 \texttt{ m/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
