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A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5
- a)172 m/s
- b)124 m/s
- c)99 m/s
- d)86 m/s
Correct answer is option 'A'. Can you explain this answer?
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A circular road of radius 1000 m has banking angle 45o. The maximum sa...
Given data:
- Radius of circular road (r) = 1000 m
- Banking angle (θ) = 45°
- Mass of car (m) = 2000 kg
- Coefficient of friction between tyre and road (μ) = 0.5
To find: Maximum safe speed of the car
1. Normal Force (N):
The normal force acting on the car can be determined by resolving the weight of the car into its components:
- Vertical component = mg * cos(θ)
- Horizontal component = mg * sin(θ)
Since the car is in equilibrium on the inclined plane, the vertical component of the weight is balanced by the normal force. Therefore, N = mg * cos(θ).
2. Centripetal Force (F_c):
The centripetal force required to keep the car moving in a circle can be calculated using the equation: F_c = m * v^2 / r, where v is the velocity of the car.
3. Frictional Force (F_friction):
The frictional force acting on the car is given by the equation: F_friction = μ * N.
4. Equating the forces:
Since the car is in equilibrium on the inclined plane, the frictional force provides the necessary centripetal force. Therefore, F_friction = F_c.
5. Substituting the values:
Using the equations F_c = m * v^2 / r and F_friction = μ * N, we can substitute the known values to solve for the maximum safe speed.
m * v^2 / r = μ * N
Substituting N = mg * cos(θ):
m * v^2 / r = μ * mg * cos(θ)
Simplifying the equation:
v^2 = μ * g * r * cos(θ)
v = √(μ * g * r * cos(θ))
v = √(0.5 * 9.8 * 1000 * cos(45°))
v = √(0.5 * 9.8 * 1000 * (1/√2))
v = √(0.5 * 9.8 * 1000 * 0.7071)
v ≈ 172 m/s
Therefore, the maximum safe speed of the car is approximately 172 m/s.
- Radius of circular road (r) = 1000 m
- Banking angle (θ) = 45°
- Mass of car (m) = 2000 kg
- Coefficient of friction between tyre and road (μ) = 0.5
To find: Maximum safe speed of the car
1. Normal Force (N):
The normal force acting on the car can be determined by resolving the weight of the car into its components:
- Vertical component = mg * cos(θ)
- Horizontal component = mg * sin(θ)
Since the car is in equilibrium on the inclined plane, the vertical component of the weight is balanced by the normal force. Therefore, N = mg * cos(θ).
2. Centripetal Force (F_c):
The centripetal force required to keep the car moving in a circle can be calculated using the equation: F_c = m * v^2 / r, where v is the velocity of the car.
3. Frictional Force (F_friction):
The frictional force acting on the car is given by the equation: F_friction = μ * N.
4. Equating the forces:
Since the car is in equilibrium on the inclined plane, the frictional force provides the necessary centripetal force. Therefore, F_friction = F_c.
5. Substituting the values:
Using the equations F_c = m * v^2 / r and F_friction = μ * N, we can substitute the known values to solve for the maximum safe speed.
m * v^2 / r = μ * N
Substituting N = mg * cos(θ):
m * v^2 / r = μ * mg * cos(θ)
Simplifying the equation:
v^2 = μ * g * r * cos(θ)
v = √(μ * g * r * cos(θ))
v = √(0.5 * 9.8 * 1000 * cos(45°))
v = √(0.5 * 9.8 * 1000 * (1/√2))
v = √(0.5 * 9.8 * 1000 * 0.7071)
v ≈ 172 m/s
Therefore, the maximum safe speed of the car is approximately 172 m/s.
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A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer?
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A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer?.
A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A circular road of radius 1000 m has banking angle 45o. The maximum safe speed of a car having mass 2000 kg will be, if the coefficient of friction between tyre and road is 0.5a)172 m/sb)124 m/sc)99 m/sd)86 m/sCorrect answer is option 'A'. Can you explain this answer?.
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