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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?
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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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Most Upvoted Answer
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?
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