Question

A car traveling at a steady 20 m/s rounds an 80-m radius horizontal unbanked curve with the tires on the verge of slipping. what is the maximum speed with which this car can round a second unbanked curve of radius 320 m if the coefficient of static friction between the car's tires and the road surface is the same in both cases?

Answer 1

Refer to the diagram shown below.

r = the radius of the curve
m =  the mass of the car
μ = the coefficient of kinetic friction
N = normal reaction

When rounding the curve, the centripetal acceleration is
$a = \frac{v^{2}}{r}$

The inertial force tending to make the car skid is balanced by the frictional force, therefore
$\mu mg = m \frac{v^{2}}{r} \\ \mu = \frac{v^{2}}{rg}$

When r = 80 m and v = 20 m/s, obtain
$\mu = \frac{(20 m/s)^{2}}{(80 \, m)*(9.8 \, m/s^{2})} = 0.51$

When r = 320 m and μ remains the same, obtain
$\frac{(v \, m/s)^{2}}{(320 \, m)*(9.8 \, m/s^{2})} =0.51 \\\\ v^{2} = 1.6 \times 10^{3} \\\\ v = 40 \, m/s$

Answer: 40 m/s

Answer 2

The maximum speed with which this car can round a second unbanked curve is 40 m/s

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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:

initial speed of the car = v₁ = 20 m/s

radius of first curve = R₁ = 80 m

radius of second curve = R₂ = 320 m

Asked:

final speed of the car = v₂ = ?

Solution:

Firstly , we will derive the formula to calculate the maximum speed of the car:

$\Sigma F = ma$

$f = m \frac{v^2}{R}$

$\mu N = m \frac{v^2}{R}$

$\mu m g = m \frac{v^2}{R}$

$\mu g = \frac{v^2}{R}$

$v^2 = \mu g R$

$\boxed {v = \sqrt { \mu g R } }$

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Next , we will compare the maximum speed of the car on the first curve and on the second curve:

$v_1 : v_2 = \sqrt { \mu g R_1 } : \sqrt { \mu g R_2 }$

$v_1 : v_2 = \sqrt { R_1 } : \sqrt { R_2 }$

$20 : v_2 = \sqrt { 80 } : \sqrt { 320 }$

$20 : v_2 = 1 : 2$

$v_2 = 2(20)$

$\boxed {v_2 = 40 \texttt{ m/s}}$

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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