Answer :
The centripetal force tending to take the car off track is
[(600 kg)(24.5 m/s)^2]/(110 m) = 3274.09 N
Let the banking angle be x.
Then the restorative force on the car is
(600 kg)(9.8 m/s^2)sin(x) = 5880sin(x) N
For dynamic equilibrium,
5880sin(x) = 3274.09
sin(x) = 0.5568
x = arcsin(0.5568) = 33.8°
Answer: 33.8°
[(600 kg)(24.5 m/s)^2]/(110 m) = 3274.09 N
Let the banking angle be x.
Then the restorative force on the car is
(600 kg)(9.8 m/s^2)sin(x) = 5880sin(x) N
For dynamic equilibrium,
5880sin(x) = 3274.09
sin(x) = 0.5568
x = arcsin(0.5568) = 33.8°
Answer: 33.8°
The appropriate banking angle is about 29.1°
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Further explanation
Centripetal Acceleration of circular motion could be calculated using following formula:
[tex]\large {\boxed {a_s = v^2 / R} }[/tex]
a = centripetal acceleration ( m/s² )
v = velocity ( m/s )
R = radius of circle ( m )
Let us now tackle the problem!
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Given:
mass of car = m = 600 kg
radius of banked curve = R = 110 m
speed of car = v = 24.5 m/s
Asked:
banking angle = θ = ?
Solution:
Firstly , we will draw free body diagram of the car as shown in the attachment.
Next , we could use Newton's Law of Motion and Centripetal Force to solve the problem as follows:
[tex]\Sigma F_y = ma_y[/tex]
[tex]N\cos \theta - mg = m(0)[/tex]
[tex]N\cos \theta - mg = 0[/tex]
[tex]N\cos \theta = mg[/tex]
[tex]\boxed{ N = mg \div \cos \theta }[/tex] → Equation 1
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[tex]\Sigma F_x = ma_x[/tex]
[tex]N \sin \theta = m \frac{v^2}{R}[/tex]
[tex]( mg \div \cos \theta ) \sin \theta = m \frac{v^2}{R}[/tex] ← Equation 1
[tex]mg \tan \theta = m \frac{v^2}{R}[/tex]
[tex]g \tan \theta = \frac{v^2}{R}[/tex]
[tex]\tan \theta = \frac{v^2}{gR}[/tex]
[tex]\theta = \tan^{-1} ( \frac{v^2}{gR} )[/tex]
[tex]\theta = \tan^{-1} ( \frac{24.5^2}{ 9.8 \times 110 } )[/tex]
[tex]\boxed{ \theta \approx 29.1^o }[/tex]
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Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
- Uniform Circular Motion : https://brainly.com/question/2562955
- Trajectory Motion : https://brainly.com/question/8656387
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Answer details
Grade: High School
Subject: Physics
Chapter: Circular Motion
