Question

Cyclist competes in a one-lap race around a flat, circular course of radius 140 m . starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60 s . the mass of the bicycle (including the rider) is 76 kg . what is the magnitude of the net force fnet acting on the bicycle as it crosses the finish line?

Answer 1

Refer to the diagram shown below.

The angular distance traveled in one lap is θ = 2π radians.

Let  α =  the angular acceleration, rad/s².
Because 1 lap was completed in t = 60 s, therefore the angular acceleration is given by
θ = (1/2)*α*t²
That is,
2π rad = 0.5*(α rad/s²)*(60 s)² 
2π = 1800 α
α = 3.49 x 10⁻³ rad/s²

The angular velocity at the end of the lap is
ω = αt
    = (3.49 x 10⁻³ rad/s²)*(60 s)
    = 0.2094 rad/s

The tangential velocity is
v = rω = (140 m)*(0.2094) = 29.32 m/s

The centripetal force acting on the cyclist at the finish line is
F = m*r*ω²
   = (76 kg)*(140 m)*(0.2094 rad/s)²
   = 466.5 N

Answer: 466.5 N

Answer 2

The magnitude of the net of force is about 470 Newton

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

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:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

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:

$s = ut + \frac{1}{2}at^2$

$2\pi R = 0(T) + \frac{1}{2}a(T)^2$

$2\pi (140) = 0 + \frac{1}{2}a(60)^2$

$280\pi = 1800a$

$a = 280 \pi \div 1800$

$a = \frac{7}{45} \pi \texttt{ m/s}^2$

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Next we will find the centripetal acceleration of the cyclist as it crosses the finish line:

$a_c = v^2 \div R$

$a_c = ( u + aT )^2 \div R$

$a_c = ( 0 + \frac{7}{45} \pi(60))^2 \div 140$

$a_c = \frac{28}{45} \pi^2 \texttt{ m/s}^2$

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Finally we could calculate the magnitude of the net force by using Newton's 2nd Law Of Motion as follows:

$\Sigma F = m\sqrt{a^2 + a_c^2}$

$\Sigma F = 76 \sqrt{(\frac{7}{45}\pi)^2+(\frac{28}{45}\pi^2)^2}$

$\Sigma F \approx 470 \texttt{ Newton}$

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Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion