Question

obtain the expressions for speed of a vehicle at banked road of friction
​

Answer 1

Answer:

Complete step-by-step answer:

Several formulas used in our calculations are-

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.

Let’s take inclination angle A of road with respect to the x-axis as shown in the above diagram. Now we can balance all the x-axis and y-axis factors of forces. Refer to the figure.

(1) 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-

NcosA=fsinA+mg⇒NcosA=μNsinA+mg⇒N(cosA−μsinA)=mg (2)

(3) Frictional force-f=μN

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)(1μtanA)

This is an expression for maximum speed of a car on a banked road in circular motion.

I hope it's helpful for

Answer 2

f → force of friction N →reaction force which will provide centripetal force. Resolve it into two components – Ncos θ and Nsin θ Also resolve friction force f into fsin θ and fcos θ. (centripetal force) mv2/R = fcos θ + Nsin θ (1) (acting towards centre) Ncos θ = mg + fsin θ or mg = Ncos θ – fsin θ Dividing eq. (1) by (2)