The magnitude of the net of force is about 470 Newton
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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:
period of the circular motion = T = 60 s
mass of the the bicycle = m = 76 kg
radius of the circuit = R = 140 m
Unknown:
magnitude of the net force = ΣF = ?
Solution:
We will use this following formula to find the tangential acceleration of the cyclist:
[tex]s = ut + \frac{1}{2}at^2[/tex]
[tex]2\pi R = 0(T) + \frac{1}{2}a(T)^2[/tex]
[tex]2\pi (140) = 0 + \frac{1}{2}a(60)^2[/tex]
[tex]280\pi = 1800a[/tex]
[tex]a = 280 \pi \div 1800[/tex]
[tex]a = \frac{7}{45} \pi \texttt{ m/s}^2[/tex]
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Next we will find the centripetal acceleration of the cyclist as it crosses the finish line:
[tex]a_c = v^2 \div R[/tex]
[tex]a_c = ( u + aT )^2 \div R[/tex]
[tex]a_c = ( 0 + \frac{7}{45} \pi(60))^2 \div 140[/tex]
[tex]a_c = \frac{28}{45} \pi^2 \texttt{ m/s}^2[/tex]
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Finally we could calculate the magnitude of the net force by using Newton's 2nd Law Of Motion as follows:
[tex]\Sigma F = m\sqrt{a^2 + a_c^2}[/tex]
[tex]\Sigma F = 76 \sqrt{(\frac{7}{45}\pi)^2+(\frac{28}{45}\pi^2)^2}[/tex]
[tex]\Sigma F \approx 470 \texttt{ Newton}[/tex]
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Grade: High School
Subject: Physics
Chapter: Circular Motion
