# Car on a frictionless banked curve

A car is on a banked curve, following a path which is part of a circle with radius $R$. The curve is banked at angle $\theta$ with the horizontal, and is a frictionless surface. What is the speed the car must go to accomplish this?

What I don't understand about this problem is why we assume there is only the normal force and the gravitational force on the vehicle. From that point onwards, I have no trouble following the solution.

The way I see it, if we're considering only that there exists a normal force and a gravitational force, something the car is doing (accelerating?) must be "adding" to the normal force. Or is it possible that the car could just be coasting, and it could keep a constant height along the banked curve?

• While the question bears a certain resemblance to a no-effort homework question (especially in formatting) a close reading shows that the OP is asking a good conceptual question quite appropriate for Physics.SE. – dmckee Dec 15 '13 at 19:53
• Yeah, that's my take on it as well. – David Z Dec 15 '13 at 21:53
• Given that the track surface is given as "friction-less", what else can the car do but coast? No steering, braking or engine thrust. – DJohnM Dec 16 '13 at 0:07

You are right in thinking that the car's acceleration is what keeps it in place, but it is important to remember that an object moving at a constant speed in a circle is accelerating (despite not speeding up). The reason for this is that acceleration is defined as a "change in velocity," and velocity is a vector quantity (i.e. it has magnitude and direction). Therefore, the car's centripetal acceleration must have a force corresponding to it through Newton's Second Law (that force is the radial component of the normal force). And by requiring the position of the car to behave in certain ways, we can calculate that force exactly.

So yes, the acceleration is "adding" to the normal force. The sloped track must apply a greater force in the radial direction than it would if the car were not moving because in that situation the car would begin moving along the radial direction (sliding down the ramp). Likewise, it is not moving in the vertical direction (sliding down the ramp) so the normal force in that direction must be greater as well (to exactly cancel its weight). Since the normal force is the vector sum of its radial and vertical components, you can determine the increased normal force.

The truth is that by stating the problem this way, we have required certain constraints on the motion of the car from which we can deduce the forces acting on it by using Newton's Second Law.

• Maybe you can clarify just a bit further? Suppose the car starts on the track an initial speed $v_0$ along the direction of the track. As @User58220 has mentioned, the car is not able to influence movements as the question is put anyways. The system of the car and the track now has no net force applied to it. Since its the acceleration of the car (not a change in speed though!), would that imply that the car would now slip downwards along the banked curve? – Ryder Bergerud Dec 16 '13 at 5:36
• Don't think of the car and the track as the system; think of just the car. The car by itself does have forces acting on it (the car-track does too but it isn't very enlightening). The car can't influence its motion once its started moving but the track can. Since the car is constrained to move along the track, if the car is moving fast enough the normal force of the track on the car will make it move in a circle. Given just the right $v_0$, the vertical component of the normal force will cancel its weight (so that sum of forces equals 0 and there is no vertical acceleration). – Geoffrey Dec 16 '13 at 6:40
• Likewise, the radial component of the normal force will exactly equal $\frac{v^2}{r}$ and imply acceleration along the radial direction. Radial acceleration of this sort corresponds to a constant and continuous change in velocity along the radial direction which results in the car moving in a circle. If $v_0$ is not just right, then the situation is more complicated and there could be some acceleration that makes the path the car follows not a perfect circle (i.e. it slides up or down the track). – Geoffrey Dec 16 '13 at 6:40
• Ok, I see now how a correct choice of $v_0$ implies the normal force and hence $a_rad$ sufficient to give us constant speed around the track for the rest of time. Considering the position as the solution perhaps of some differential equation, it is periodic. Would it perhaps be unstable, (or a repeller if all trajectories of the vehicle that didn't start with v_0 would diverge from the path of the periodic solution). Would looking at the total energy of the car be the right way to approach this? That is $v^2_{t2} = (h_1 - h_2)g -v^2{t1}$ – Ryder Bergerud Dec 16 '13 at 7:59

See the diagram for guidance : http://picpaste.com/p012-Lrn6pmdn.jpg

Draw the gravity vector and the centripetal vector, then the resultant of the two, the banking should be normal to this resultant for no side forces.

m = car mass in kg
g = local gravity rate in (m/s)/s.
v = velocity in m/s.
r = radius to centre of gravity in metres.
A = banking angle in degrees.

$A = \tan^{-1}(\frac{v^2}{rg})$

(m's cancel out)

• Your link does not appear to work. At least it doesn't lead anywhere for me. – Geoffrey Dec 28 '13 at 3:26
• Use imgur.com for linking pics – ja72 Dec 28 '13 at 8:09