Question

15. Write the expression for maximum velocity of a car on a circular level road and explain the terms.
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Answer 1

Answer:

Balancing all the forces acting along the x-axis-

fcos(A)+Nsin(A)=mv2r

Substituting the value f=μN into equation, we get

μNcos(A)+Nsin(A)=mv2r⇒N(μcosA+sinA)=mv2r(1)

(2) Balancing all the forces acting along the y-axis-Gravitational attraction force – mg

Frictional force (f) – acting between car’s tire and road, and parallel to the slope of road

Normal force (N)- It is a reaction force of the gravitational force exerted by the road on car

Centripetal force (mv2r) - acting towards the center of the circular path followed by the car.Frictional force-

f=μN

NcosA=fsinA+mg⇒NcosA=μNsinA+mg⇒N(cosA−μsinA)=mg

(2)Gravitational attraction force – mg

Frictional force (f) – acting between car’s tire and road, and parallel to the slope of road

Normal force (N)- It is a reaction force of the gravitational force exerted by the road on car

Now we will use these three equations to find the value of speed (v).

Now we can divide equation (1) and (2), we get

v2rg=(μ+tanA)(1−μtanA)⇒v=rg(μ+tanA)(1−μtanA√

This is an expression for maximum speed