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A cyclist turns around a curve at 15 miles/hour. If he turns at double the speed, the tendency to overturn is
- a)Doubled
- b)Quadrupled
- c)Halved
- d)Unchanged
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A cyclist turns around a curve at 15 miles/hour. If he turns at double...
Explanation:
When a cyclist turns around a curve, there are two types of forces acting on the cyclist and the bicycle. These two forces are:
1. Centripetal force: This force acts towards the center of the curve and is responsible for keeping the cyclist and the bicycle moving in a circular path.
2. Centrifugal force: This force acts in the outward direction and is responsible for the tendency of the cyclist and the bicycle to overturn or skid off the curve.
The relationship between the speed of the cyclist and the centrifugal force exerted on him is given by the formula:
F = mv^2 / r
where F is the centrifugal force, m is the mass of the cyclist and the bicycle, v is the speed of the cyclist, and r is the radius of the curve.
Now, let's consider the two scenarios given in the question:
Scenario 1: The cyclist turns around the curve at 15 miles/hour.
Scenario 2: The cyclist turns around the same curve at double the speed, i.e., 30 miles/hour.
In scenario 1, the centrifugal force exerted on the cyclist is:
F1 = mv1^2 / r
where v1 = 15 miles/hour
In scenario 2, the centrifugal force exerted on the cyclist is:
F2 = mv2^2 / r
where v2 = 30 miles/hour
To find the tendency to overturn, we need to compare the two forces, i.e., F2/F1.
F2/F1 = (mv2^2 / r) / (mv1^2 / r)
= (v2/v1)^2
= (30/15)^2
= 4
Therefore, the tendency to overturn is quadrupled when the cyclist turns around the curve at double the speed.
Hence, the correct answer is option B, i.e., quadrupled.
When a cyclist turns around a curve, there are two types of forces acting on the cyclist and the bicycle. These two forces are:
1. Centripetal force: This force acts towards the center of the curve and is responsible for keeping the cyclist and the bicycle moving in a circular path.
2. Centrifugal force: This force acts in the outward direction and is responsible for the tendency of the cyclist and the bicycle to overturn or skid off the curve.
The relationship between the speed of the cyclist and the centrifugal force exerted on him is given by the formula:
F = mv^2 / r
where F is the centrifugal force, m is the mass of the cyclist and the bicycle, v is the speed of the cyclist, and r is the radius of the curve.
Now, let's consider the two scenarios given in the question:
Scenario 1: The cyclist turns around the curve at 15 miles/hour.
Scenario 2: The cyclist turns around the same curve at double the speed, i.e., 30 miles/hour.
In scenario 1, the centrifugal force exerted on the cyclist is:
F1 = mv1^2 / r
where v1 = 15 miles/hour
In scenario 2, the centrifugal force exerted on the cyclist is:
F2 = mv2^2 / r
where v2 = 30 miles/hour
To find the tendency to overturn, we need to compare the two forces, i.e., F2/F1.
F2/F1 = (mv2^2 / r) / (mv1^2 / r)
= (v2/v1)^2
= (30/15)^2
= 4
Therefore, the tendency to overturn is quadrupled when the cyclist turns around the curve at double the speed.
Hence, the correct answer is option B, i.e., quadrupled.
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A cyclist turns around a curve at 15 miles/hour. If he turns at double...
This is because μ is directly proportional to v².
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