Answer:
18.5 m/s
Explanation:
On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:
[tex]\mu mg = m\frac{v^2}{r}[/tex]
where
[tex]\mu[/tex] is the coefficient of static friction between the tires and the road
m is the mass of the car
g is the gravitational acceleration
v is the speed of the car
r is the radius of the curve
Re-arranging the equation,
[tex]v=\sqrt{\mu gr}[/tex]
And by substituting the data of the problem, we find the speed at which the car begins to skid:
[tex]v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s[/tex]
The car will begin to skid sideways at 18.5 m/s
[tex]\texttt{ }[/tex]
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 )
[tex]\texttt{ }[/tex]
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 !
[tex]\texttt{ }[/tex]
Given:
mass of car = m = 1000 kg
radius of curve = R = 100 m
coefficient of static friction = μ = 0.350
Asked:
speed of the car = v = ?
Solution:
We will derive the formula to calculate the maximum speed of the car:
[tex]\Sigma F = ma[/tex]
[tex]f = m \frac{v^2}{R}[/tex]
[tex]\mu N = m \frac{v^2}{R}[/tex]
[tex]\mu m g = m \frac{v^2}{R}[/tex]
[tex]\mu g = \frac{v^2}{R}[/tex]
[tex]v^2 = \mu g R[/tex]
[tex]\boxed {v = \sqrt { \mu g R } }[/tex]
[tex]v = \sqrt { 0.350 \times 9.8 \times 100 }[/tex]
[tex]v = \sqrt { 343 }[/tex]
[tex]v = 7 \sqrt{7} \texttt{ m/s}[/tex]
[tex]\boxed {v \approx 18.5 \texttt{ m/s}}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Circular Motion
