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Forces on a car in circular motion

  1. May 20, 2014 #1
    I've been trying to work out the forces on a car in circular motion around a turn and I'm having trouble understanding what causes a friction force to be directed inward toward the center of the circle.

    I understand that on a straight path, the torque on the wheels causes them to push back against the ground and the opposing force of the ground pushing on the car moves it forward. Ignoring air resistance, this thrust would have to be equal to the friction force between the tires and the ground to keep the car moving at a constant velocity.

    What happens at that instant where the wheel is turned and the car enters a circular path. Based on the straight path case, by applying a torque to turn the wheels, the thrust is redirected at a slight angle to its original direction. I would assume for the speed to remain constant, the thrust and friction will still be balanced in the new direction of motion. So what gives rise to a net friction force directed toward the center of the circular path?

    Isn't it the torque applied to turn the wheels that is causing the circular motion? After all if the driver lets go of the wheel the car will return to a straight line path.
     
  2. jcsd
  3. May 20, 2014 #2

    Nugatory

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    You are misunderstanding the relationship between thrust and friction here. The motor's thrust is applied to the axle, which in turn tries to slide the tire's contact patch (the bit that touches the ground) backwards. The frictional force between the ground and the contact patch resists the this backwards motion so points forward; it is this forward force from friction that drives the car forward against air resistance to maintain a constant speed. But with that said....

    If the frictional force at the tire's contact path remained constant, the car would have to slow down as it enters a turn, for pretty much the reason that you describe: The force vector through the tire's contact patch points slightly sideways relative to the direction of travel of the car so its component in the direction of travel is no longer sufficient to overcome air resistance and rolling resistance. Race drivers even have a term for this effect - "scrubbing off speed" refers to the way a car can be slowed by taking it through a turn without applying more power. However, in normal driving the frictional force usually can be increased because the driver can apply a tad more power to maintain his speed through the turn.... scrubbing off speed happens when more power cannot be applied because the tire is already at its limit of adhesion and any more power would just cause it to start sliding.

    The steering system of a car is designed so that it will tend to return to straight if you stop pushing on the steering wheel; if it weren't for this you could turn the wheel, then let go of it and the wheel would stay put and the car would keep turning.
     
  4. May 20, 2014 #3

    UltrafastPED

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    The rolling tire has a segment in contact with the road - in the idealized case it is a point, but real tires have a contact segment. The portion of the tire in contact is stationary wrt the roadway; you move forward because the wheel is turning, changing the contact patch.

    When tou are going around a bend there are additional forces; at the tires there is a centrifugal force pushing the tire away from the center of curvature (center of the bend); if your car stays on the road it is because of a sideways frictional force opposing the inertial forces of the turn. If you watch a movie you can see the tire flex or deform sideways.
     
  5. May 21, 2014 #4
    Sorry, I probably could have explained that better; I do understand the relationship between the thrust and friction up to this point. The difficulty I'm having is working out the force vectors once the car enters a turn.

    I think I may understand now though. At the instant the tires are turned and the car begins to move into the circular path, the forward force of friction from the contact patch would be applied at the new angle, but the car would still want to move in a straight line. So, the centripetal force is the component of the forward force that is perpendicular to the original direction of motion. While the parallel component is overcome by the drag forces in that direction; slowing down the car in its original direction and forcing it onto the new path. Does that sound correct or am I still missing something?
     
  6. May 21, 2014 #5

    rcgldr

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    The longitudinal forces on a tire are related to torque from the engine or brakes, and the rolling resistance (which is different than friction). The lateral forces on a tire are related to centripetal acceleration or lean angle of the pavement. The end result of steering a car is a centripetal acceleration because the tires will tend to roll and generate centripetal acceleration rather than just skid while going straight (assuming friction is zero or nearly zero).

    If the axis of the front and rear tires are extended, the point at where those axis lines cross would be the idealized center of the circular path of the car, except that the tires will deform somewhat due to the centripetal related load (slip angle) so the actual radius will be slightly larger than the idealized radius. This assumes an idealized Ackerman steering setup, where the inner tire is steered more than the outer tire so that the extended axis of all 4 tires always cross at a single point. Without Ackerman steering, the extended axis cross at two points, one for the inner front tire, the other for the outer front tire, and the effective radius will be somewhere between those two points (plus any effect related to tire deformation).
     
    Last edited: May 21, 2014
  7. May 21, 2014 #6
    The attached relates to the question, and may be helpful.

    Cheers,
    Terry
     

    Attached Files:

  8. May 23, 2014 #7
    I've been pretty busy so I haven't had time to more than skim the article, but the information looks really useful, thank you. I'll have to see if I can find the other parts of the method.
     
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