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Circular Motion of Car in Vertical Bridge

  1. Jun 22, 2007 #1
    1. A car of mass 1500kg moving at constant speed of 80km/hr passes over a humpback of a bridge of radius of curvature r. The car remains in contact wit the road as it passes over the top.

    a) Express N, net force exerted by the car on the road, in terms of r.

    b) Determine the minimum curvature required of the car to just remain in contact with the road.

    Generally i do understand the basic concept behind the questions, where [Weight of Car - Normal Reaction of the road on the Car] = Result Force = Centripetal Force

    But why is that the case? Why is it that Mg is greater than NR? If it is different, we can see Mg is not the force exerted on the road by the car. So what is the force exerted by the car on the road that causes the Normal Reaction on the car by the road. Is it a component of the weight? Where does the force come from?
  2. jcsd
  3. Jun 22, 2007 #2


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    In order to go around a curve there must be some force that pushes (pulls) the object around it. The only force that can do it in this case is its weight. Due to inertia the car wants to continue in a straight line though, that is it wants to continue moving in a straight line tangentially to the curvature of the bridge. This reduces the force that it pushes down onto the surface of the bridge - the normal force. If the speed of the car increases it will eventually move faster than it falls down and it will become airborn - a projectile in fact.
  4. Jun 22, 2007 #3
    Does your reasoning apply for cars at the top of the bridge? It seems like for the question, NR is still less than the weight. So its due to inertia? Coz at the top of the motion, the forces act like on a straight line motion, mg is vertical, so shouldn't NR be = to Mg coz its like on the straight road.

    And also, for a car on the bridge not at the top, Centripetal Force = (mg)(cos@) - NR ?? Where @ is the angle the weight makes to the Normal.
  5. Jun 22, 2007 #4

    Doc Al

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    The difference here is that the car going over the top of the bridge is accelerating--there is a downward acceleration due to the circular motion. On a straight road, the vertical acceleration is zero so N = mg.

    Sounds good.
  6. Jun 22, 2007 #5
    Oh Thank the both of you guys so much.

    Though how exactly inertia reduces the NR force explained in terms of forces (makes me think if inertia is a force), i am still not very sure. But the idea is intuitive and i appreciate you guys explaining it for me.

    Another question, if there is a horizontal circular disk with attached small blades at the sides, it is said that if the rotation is more then 500 rad per s then the blades will break and fly off. My teacher explained wit is that the force that holds the blades to the wheel may not be large enough to provide the centripetal force required. So the blades provide centripetal force?

    What actually provides centripetal force? Centripetal force is the force needed to maintain a circular motion with constant speed right? It is not the force that starts of the motion right? To start of a circular motion from rest, you need to have a force applied along the lines of the speed as well as the centripetal force?

    And i know this is a bit of the main topic here but, How do cars actually move?
    Coz my teacher said that the friction provides the reaction force to the force exerted by the car on the road. This reaction force is equal and opposite in direction and it is exerted on the car. This force allows the car to move.
    Then i thought about Fapplied - friction = ma, if friction moves the car, then whats the concept behind Fapplied - friction = Fresultant = ma? What did my teacher mean when he said, when friction is not enough, the car cannot move. So the friction actually is not the reaction force to the exerted force by the car?

    Sorry for the number of questions i asked. Exam soon. =[
  7. Jun 22, 2007 #6

    Doc Al

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    In order for anything to move in a circle, something must be exerting a centripetal force on it. Imagine you are holding a rock in your hand and then you spin around in a circle. Your hand is providing the centripetal force on the rock. If you spin fast enough, the amount of force needed to maintain the rock's circular motion will be greater than your hand can provide--that's when it flys out of your hand.

    In the case of the blades attached to the disk, it is the disk that pulls the blades inward--the disk provides the centripetal force. The attachment between blade and disk is only so strong--at some point the blades break off if the disk spins too fast.

    Right. To get something going in a circle in the first place requires some force tangent to the circle.

    In order for the car to accelerate, something must provide an external force. That something is the road. Imagine being on a patch of frictionless ice--you can step on the gas all you want: No friction from the road means no acceleration of the car.

    And like any other force, there's always an "action-reaction" pair involved. In order for the road to push on the car, the car must push on the road.

    What you really need is Newton's 2nd law. The net force on an object = ma. In the case of the car on a horizontal stretch of road, the only horizontal force on it (ignoring air resistance) is the friction between tires and road. So: Friction force = ma. (Friction can act in the direction of motion, as when increasing speed; or opposite to the motion, as when applying the brakes.)

    That "Fapplied - friction" stuff would apply like this: Say you were dragging some crate across the floor using some applied horizontal force Fa. If friction acts to resist the motion, then the net force on the crate is Fa - friction, which must then equal "ma".

    Make sense?
  8. Jun 22, 2007 #7

    Thank you very much for the above answers, but regarding the cars.

    So friction = ma, because friction is the only horizontal force. Since friction is = to Fapplied, coz it is its reaction, and there are no other opposing force on this friction, it becomes the resultant force. Thus, as soon as the car starts, it already starts to accelerate? Then how do you get a constant speed on a car? There is only a resultant force with no opposition. It is static rolling friction that applies this force onto the car right?

    When the car exerts force larger then the max static friction, then what happens to the extra force?
  9. Jun 22, 2007 #8

    Doc Al

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    I'm not exactly sure what you are saying about Fapplied and that it being a reaction. If you want to think of friction as being an applied force, no problem. But the reaction to friction on the tires is the force the tires exert on the road.

    Nonetheless, ignoring air resistance, the force of friction that the road exerts on the tires is the only (horizontal) force and thus the net force on the car.

    In order to start moving, it must accelerate. I hope that's obvious! And that requires a force.
    In order to get a constant speed in a straight line you need to have a net force of zero. If there were no internal resistance tending to slow the wheels, no air resistance, and no rolling friction, then you could just coast along. But in the "real world" you need to apply the gas to keep the wheels spinning at the right speed.

    Not sure what you mean. It's physically impossible for the car to exert a force on the road greater than the maximum static friction.

    Perhaps you mean: What happens if, by stepping on the gas, we exert more force on the wheels than can be balanced by road friction? In that case, the wheels start spinning faster, losing their grip on the road. (Friction changes from static to the weaker kinetic, which means you have less force available to control the car.)
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