Question

"a car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.45 m/s2. the car makes it one quarter of the way around the circle before it skids off the track. from these data, determine the coefficient of static friction between the car and track"

Answer 1

Answer:

0.148

Explanation:

For a car travelling on a flat curve, the centripetal force is provided by the frictional force between the tires and the road:

$m\frac{v^2}{r}=\mu mg$

where

m is the car's mass

v is the tangential speed

r is the radius of the curve

$\mu$ is the coefficient of static friction

g is the gravitational acceleration

Dividing by m (the mass), we obtain an expression for the centripetal acceleration:

$a_c = \frac{v^2}{r}=\mu g$

Therefore, by substituting a = 1.45 m/s^2 and g =9.8 m/s^2, we can solve the formula to find the coefficient of static friction:

$\mu=\frac{a_c}{g}=\frac{1.45 m/s^2}{9.8 m/s^2}=0.148$

Answer 2

The coefficient of static friction between the car and track is about 0.465

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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:

tangential acceleration = a_t = 1.45 m/s²

initial angular velocity = ωo = 0 rad/s

position angle = θ = ¹/₄ rev = ¹/₂π rad

Asked:

coefficient of static friction = μ = ?

Solution:

Firstly, let's find the final angular velocity by using following formula:

$\omega^2 = \omega_o^2 + 2\alpha \theta$

$\omega^2 = \omega_o^2 + 2(\frac{a_t}{R}) \theta$

$\omega^2 = 0 + 2(\frac{1.45}{R}) (\frac{1}{2}\pi})$

$\omega^2 = \frac{1.45\pi}{R}$

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Next , we could use the centripetal acceleration as follows:

$\Sigma F = ma$

$f = ma$

$\mu N = m \omega^2 R$

$\mu mg = m \omega^2 R$

$\mu g = \omega^2 R$

$\mu = \omega^2 R \div g$

$\mu = \frac{1.45\pi}{R} R \div g$

$\mu = 1.45 \pi \div g$

$\mu = 1.45 \pi \div 9.8$

$\mu \approx 0.465$

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Conclusion:

The coefficient of static friction between the car and track is about 0.465

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant