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The coefficient of frictionbetween the tyres and the road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)
- a)√15 m s-1
- b)√3 m s-1
- c)√30 m s-1
- d)√10 m s-1
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
The coefficient of frictionbetween the tyres andthe road is 0.1. The m...
Here, r = 3m, μ = 0.1, g = 10ms−2
The maximum speed with which a cyclist can take a turn without skidding is
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The maximum speed with which a cyclist can take a turn without skidding is
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The coefficient of frictionbetween the tyres andthe road is 0.1. The m...
To find the maximum speed with which a cyclist can take a circular turn without skidding, we need to consider the centripetal force and the maximum frictional force.
The centripetal force required to keep the cyclist moving in a circular path is given by:
Fc = m*v^2 / r
Where:
Fc = Centripetal force
m = Mass of the cyclist
v = Velocity of the cyclist
r = Radius of the circular turn
The maximum frictional force that can be exerted between the tires and the road is given by:
Ffriction = μ * N
Where:
Ffriction = Frictional force
μ = Coefficient of friction
N = Normal force (equal to the weight of the cyclist)
Since the cyclist is not skidding, the maximum frictional force is equal to the centripetal force:
Ffriction = Fc
Substituting the equations for Fc and Ffriction:
μ * N = m*v^2 / r
Since N = m*g (where g is the acceleration due to gravity), we can rewrite the equation as:
μ * m * g = m*v^2 / r
Simplifying the equation:
v^2 = μ * g * r
Taking the square root of both sides:
v = sqrt(μ * g * r)
Given that the coefficient of friction (μ) is 0.1, the acceleration due to gravity (g) is 9.8 m/s^2, and the radius (r) is 3m, we can substitute these values into the equation:
v = sqrt(0.1 * 9.8 * 3)
v = sqrt(2.94)
v ≈ 1.71 m/s
Therefore, the maximum speed with which a cyclist can take a circular turn of radius 3m without skidding is approximately 1.71 m/s.
The centripetal force required to keep the cyclist moving in a circular path is given by:
Fc = m*v^2 / r
Where:
Fc = Centripetal force
m = Mass of the cyclist
v = Velocity of the cyclist
r = Radius of the circular turn
The maximum frictional force that can be exerted between the tires and the road is given by:
Ffriction = μ * N
Where:
Ffriction = Frictional force
μ = Coefficient of friction
N = Normal force (equal to the weight of the cyclist)
Since the cyclist is not skidding, the maximum frictional force is equal to the centripetal force:
Ffriction = Fc
Substituting the equations for Fc and Ffriction:
μ * N = m*v^2 / r
Since N = m*g (where g is the acceleration due to gravity), we can rewrite the equation as:
μ * m * g = m*v^2 / r
Simplifying the equation:
v^2 = μ * g * r
Taking the square root of both sides:
v = sqrt(μ * g * r)
Given that the coefficient of friction (μ) is 0.1, the acceleration due to gravity (g) is 9.8 m/s^2, and the radius (r) is 3m, we can substitute these values into the equation:
v = sqrt(0.1 * 9.8 * 3)
v = sqrt(2.94)
v ≈ 1.71 m/s
Therefore, the maximum speed with which a cyclist can take a circular turn of radius 3m without skidding is approximately 1.71 m/s.
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The coefficient of frictionbetween the tyres andthe road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)a)√15 m s-1b)√3m s-1c)√30m s-1d)√10m s-1Correct answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The coefficient of frictionbetween the tyres andthe road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)a)√15 m s-1b)√3m s-1c)√30m s-1d)√10m s-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The coefficient of frictionbetween the tyres andthe road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)a)√15 m s-1b)√3m s-1c)√30m s-1d)√10m s-1Correct answer is option 'B'. Can you explain this answer?.
The coefficient of frictionbetween the tyres andthe road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)a)√15 m s-1b)√3m s-1c)√30m s-1d)√10m s-1Correct answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about The coefficient of frictionbetween the tyres andthe road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)a)√15 m s-1b)√3m s-1c)√30m s-1d)√10m s-1Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The coefficient of frictionbetween the tyres andthe road is 0.1. The maximum speed with which a cyclist can take a circular turn of radius 3m without skidding it.(Take g = m s-2)a)√15 m s-1b)√3m s-1c)√30m s-1d)√10m s-1Correct answer is option 'B'. Can you explain this answer?.
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