What is the minimum radius of a circle along which a cyclist can ride with a velocity 18 km/hr if the coefficient of friction between the tyres and the road is = 0.5
Answers
Answer: The radius is 5m
Explanation:18km/hr =5m/s
we know,
V = √urg
5 = √0.5×r×9.8
squaring on both sides,
25 = r×4.9
∴r is approximately 5m
Answer:
The minimal radius of the circle alongside which the bicycle owner can journey with a pace of 18 km/hr (or 5 m/s) and a coefficient of friction of 0.5 is about 13.88 meters.
Explanation:
From the above question,
The minimal radius of a circle alongside which a bicycle owner can journey with a pace of 18 km/hr, we want to use the equation for centripetal force:
F = m / r
where F is the centripetal force, m is the mass of the bike owner and the bicycle, v is the speed of the cyclist, and r is the radius of the circle.
The most frictional pressure that can act on the bicycle owner and the bicycle is given by:
f = μN
where f is the most frictional force, μ is the coefficient of friction between the tyres and the road, and N is the everyday pressure appearing on the bicycle owner and the bicycle.
Since the centripetal pressure required to journey in a circle is supplied by way of friction, we can equate these two forces and resolve for the radius:
F = f
m / r = μN
Substituting N = mg (where g is the acceleration due to gravity), we get:
m / r = μmg
Solving for r, we get:
r = m / (μmg)
Substituting the given values, we get:
r = (3.6 m/s) / (0.5 x 9.8 m/
)
r = 13.88 meters
Therefore,
The minimal radius of the circle alongside which the bicycle owner can journey with a pace of 18 km/hr (or 5 m/s) and a coefficient of friction of 0.5 is about 13.88 meters.
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