IIT JEE , AIEEE Forum - Physics , Mechanics - Circular Motion-3

kane

a vehicle is moving with a velocity 'v' on a curved road of width 'b' and radius of curvature 'R'. for counteracting the centrifugal force on the vehicle,the difference in elevation required in b/w the outer and the inner edges of the road is :

ans: v²b/Rg

kindly solve all the steps

rabindra8

I must solve your question if i understand the question . so plz you once again write the question clearly.

rooney

I got the answer but it is valid when the width of road is like this.

Suppose the track is banked at angle of Q. So, is the width of road the width of the base (gettin the answer here) or the width along angle Q (not getting the answer you gave).

kane

post the solution in which you are getting the answer given by me

rooney

Consider the center C. Now at distance R (assume that b<<<R), the road starts elevating. Difference in elevation is diff in height. Let angle of elevation be Q. Suppose that b is width of the road along angle Q. Balance forces
(Not considering F as no info is given)
NSin Q= mv^2/R
NCosQ = mg

tanQ = v^2/Rg

Now height of road is ~
1>Assuming that width is the length of the projection of the road on the ground)
In the triangle, h/b = tan Q
btanQ = bv²/Rg <------Answer you gave

2>But width i guess should be the length along angle Q.
So, height = bsinQ = Not equal to bv^2/Rg

Another case could be that angle Q has been assumed to be very small so, sin Q= tanQ. (but this is very unlikely)

kane

Another case could be that angle Q has been assumed to be very small so, sin Q= tanQ. (but this is very unlikely)
my sir says that this is rarely possible but possible

rabindra8

first of all i want to say that centrifugal force is applied when the frame is non inertial i.e. it rotates or goes with acc. but here road is just at rest then how this force develop . and difference in elevation means the height of the outer edge with respect to inner edge. . if i am correct then tell me if i am wrong then you simplify the question.

rabindra8

Nsinx = mv^2/R Ncosx = mg

now, tanx = v^2 /rg. then find out sinx= v^2 /

(R^2g²+v⁴)

now elevation is bsinx and this equal to == bv^2 /

(R^2g²+v⁴) which is not match with your ans. it is match with your ans when b in the given question is not along angle rather it is along base.

then tanx = h/b where h is height then h= v^2b /rg.

i think some thing is wrong in question.

rooney

@rabindra
"when the frame is non inertial i.e. it rotates or goes with acc."

Didnt get this. Can you explain again.

Force developed due to the CAR going in a circular path. How can a road go in a circular path ?

rabindra8

hi , rooney ,, you are not understand whatever i want to say. you once read the chapter circular motion from hc verma physics volume 1. then all thing is clear.

So.please read that once again. i request to you for this.

rooney

dude...when a string is rotating having a mass at its end, we consider the net effect of all the forces to be acceleration towards the center with magnitude v^2/R. So, T = mxa = mv^2/r

Another way to comprehend this is to say , that a centrifugal force away from the center along the radius is working on the mass m. Since the motion of the mass is tangential to the circle, we balance the forces
T = Centrifugal force = mv^2/R

Similarly, when the car is moving in a circular track, it is experiencing a centrifugal force of mag. mv^2/R outside OR the net acc. of car is towards the center with mag. of acc. being v^2/R.
So, where does the point of frame being inertial or inertial come into picture.

The situation can be interpreted either ways.

rabindra8

hi

rooney,

why are you rigid on your wrong concept.

now you look the last paragraph of page number 105 of my heartist book H.C.Verma.

that paragraph is written as....... It is a common misconception among the beginners that centrifugal force acts on a particle because the particle goes on a circle. Centrifugal force acts (or is assumed to act) because we describe the particle from a rotating frame which is non intertial and sitll use Newton's Laws.

Now , i think you understand the concept. and for your information i teaches some students of which are preparing for iit ,, it's O.K.

rabindra8

hi rooney

if it is clear to you then reply me.please.

rooney

In the question by kane, the centrifugal force refers to tendency of the car to be thrown outwards of the track while moving in a circular track. It is this tendency which is being considered.

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