Question

what is the smallest radius of article at which a motorcyclyst can travel if his speed is 60km/hr and the coefficient of static friction between the tires and the road is 0.32?

Answer 1

Answer:

• Smallest radius at which motorcyclist can travel = 88.6 metres

Explanation:

Given:-

• Velocity of motorcyclist, v = 60 km/h = 60 × 5/18 = 50/3 m/s
• Coefficient of static friction between the tires and road, μ = 0.32

To find:-

• Smallest radius at which a motorcyclist can travel in the same velocity of 60 km/h, r =?

Formula required:-

• Formula to calculate centripetal force

        F(centripetal) = m v² / r

• Formula to calculate the frictional force

       F(frictional) = μ m g  

[ Where m is mass of the body, v is velocity, r is the smallest radius for banking, μ is coefficient of friction, g is the acceleration due to gravity ]

Solution:-

Taking the mass of motorcyclist along with motorcycle is 'm', his velocity is 'v', coefficient of friction between tires and road is 'μ' and 'r' is the radius at which motorcyclist can travel with and 'g' as acceleration due to gravity.

• For safe ideal banking turn, When the motorcycle will be at the edge of sliding/slipping on the road.,  The centripetal force must be equal to the frictional force between the tires and road.

→ Centripetal force = Frictional force

→ m v² / r = m μ g

→ v² / r = μ g

→ v² = μ r g

→ ( 50 / 3 )² = 0.32 × r × 9.8

→ 2500 / 9 = 3.136  r

→ r = 88.577  m = 88.6  m   [approx.]

Therefore,

• The least radius for safe banking turn by the motorcycle is 88.6 metres (approximately).