Banked Curve II. Consider a wet roadway banked as in Example 5.23 (Section 5.4 ), where there is

Banked Curve II. Consider a wet roadway banked as in Example...

Key Concepts

Banked Curve Dynamics

A banked curve is designed with an inclined surface so that part of the normal force provides a centripetal component naturally, assisting vehicles in negotiating curves. This setup reduces the reliance on friction alone by balancing gravitational and inertial forces, helping maintain steady circular motion even when friction is limited.

Static Friction

Static friction is the resistive force that prevents a stationary contact from slipping, and it acts up to a maximum value determined by the coefficient of static friction and the normal force. In the context of a banked curve, static friction is critical in preventing a vehicle from sliding either up or down the incline when its speed deviates from the ideal value that the banking provides.

Centripetal Force Requirement

Centripetal force is the net inward force necessary to keep an object moving in a circular path. On a banked curve, this force is achieved through the horizontal components of the normal force and, if needed, the frictional force, ensuring that the vehicle all but remains in circular motion around the curve.

Force Decomposition

Analyzing the forces on a banked curve requires decomposing the forces, such as gravity and the normal force, into components parallel and perpendicular to the inclined surface. This decomposition is essential in solving for the conditions of equilibrium both perpendicular to the road (to prevent vertical acceleration) and parallel to the road (to prevent sliding), thus determining the restrictions on vehicle speed.

Transcript

00:02 All right.

00:03 I will start with drawing a free -by -the -diagram for the car.

00:11 So we have a car and the road is banked.

00:22 And there's an angle here which is given as beta.

00:27 And it makes a circle like this.

00:34 And since it's making a, you know, since it has a circular motion, there is going to be a radial acceleration going this way, going towards the center.

00:46 And this radial acceleration is v squared over r.

00:55 So now i will just draw all the forces exerted on the car.

01:00 So first of all, i will start with the weight, mg, which is going straight down.

01:07 And then there's the normal force that's perpendicular to the road.

01:13 And last of all, there's going to be the friction force, right? and the friction force can be either going up or going down.

01:23 So in order to determine the direction of the friction force, so in the first part, we are looking for the maximum speed, right? so for the maximum speed, the car tends to go up.

01:36 So since the car tends to go up, the friction force opposes the motion, which suggests us that the friction force will be going down for the first part.

01:53 So here is the friction force.

01:56 And i will just label the angles everywhere.

02:02 So since this angle here is beta, this is also beta.

02:10 And this angle over here, the friction force with the horizontal, it's also beta.

02:19 All right.

02:20 And i also label this axis as y and this axis is y.

02:31 And the horizontal axis here, this is x.

02:40 So the first equation that i will write is for the friction force.

02:46 So the friction force is equal to the normal force times the static friction coefficient.

02:56 So here we need to be a little bit careful.

02:58 First of all, this friction coefficient is the static one, not the kinetic.

03:03 The reason is the cart hasn't slipped yet.

03:06 That's why it is still the effective friction force is still the static friction force.

03:14 That's why we need to use static friction coefficient here instead of the kinetic one.

03:19 And here, this f, the friction force, is exactly equal to the normal force times the friction coefficient because as we are looking for the maximum speed, at the maximum speed, the friction force will reach its maximum value.

03:36 So that's why this equation here should hold.

03:41 And the second equation that i will write is the net force on the vertical axis.

03:50 And there's no acceleration of the car in the vertical axis.

03:54 So that's why this should be zero.

03:58 And the net force in the vertical axis is there is one vertical component of the normal force, which is n times cosine beta.

04:13 And this should be equal to all the vertical forces going down, which are m g, which is going straight down, plus there's the vertical component of the friction force, which is f times sine beta.

04:34 So now in this equation, i can basically, plug in the first equation here for f in terms of the normal force so that the second equation becomes n cosine beta equals m g plus for the friction force now i have n times muesthetic times sine beta yeah.

05:12 Good.

05:13 So if you look at this expression here, actually we have n here and also n here.

05:22 So basically i can solve for n.

05:25 So n is the only unknown variable here.

05:29 So that's why n can be isolated in terms of the other variables.

05:35 Then if you do that, you get mg on the right hand side.

05:39 On the left hand side, you have cosine base.

05:44 Minus mu -stallic times sine beta.

05:54 And the third equation that i will write is going to be the net force on the horizontal axis this time.

06:07 So, but this time, this won't be zero, like in the vertical case.

06:13 This is not zero because there is a horizontal acceleration, which is the radial acceleration, and that's why this is going to be m times the acceleration, which is the radial acceleration b squared over r.

06:27 If i write the horizontal forces, all the horizontal forces, i see that there is horizontal component of the normal force, which is n times sine beta.

06:44 And there's the horizontal component of the friction force as well in the same direction as the horizontal component of the normal force, that's why i added f times it is cosine equals m v squared over r.

07:05 Good.

07:06 Again, for f, i can make use of the first equation here.

07:14 For f, i can just basically plug in n times muesthetic using the first equation.

07:23 And if i do this, i will just rewrite it here.

07:30 This is just a continuation of the equation three.

07:34 So it becomes n sine beta plus for the friction force, it is nmustatic times cosine beta.

07:49 I'm just rewriting this and equals mv squared over r.

07:58 So as you see, we have n here and here, and we have v, which we are trying to solve for.

08:06 So for n, actually previously, we already found it in terms of all the given variables in the problem, right? so now i can plug in n here and also here.

08:23 If we do that, actually i will just first factor out n.

08:35 So it becomes like this.

08:47 All right.

08:47 Now i will plug in the expression of n that we found previously here.

08:58 So if i do that, it is, i'm just copying this expression of n directly here, times the term here, parentheses, sine beta plus mu static cosine beta equals m v squared over r.

09:38 So now you see that ms cancel at both sides...

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