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Uniform circular motion problem

  1. Jun 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A police officer in hot pursuit drives her car through a circular turn of radius 356 m with a constant speed of 70.9 km/h. Her mass is 57.1 kg.What are (a) the magnitude and (b) the angle (relative to vertical) of the net force of the officer on the car seat? (Hint: Consider both horizontal and vertical forces.)

    2. Relevant equations
    R=v2/a, F=mv2/R, R=V02sin(2θ0)/g

    3. The attempt at a solution
    20150626_220733_zpssecmrdc7.jpg
    Im dont see why my attempt at part (a) is wrong my answer is 62.2 N
     
  2. jcsd
  3. Jun 26, 2015 #2

    Simon Bridge

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    How did you use the hint?
    How many forces are acting on the officer?
     
  4. Jun 26, 2015 #3
    I guess i just found the horizontal force or the force pointing toward the center. There must also be a vertical force in the direction of the velocity, but velocity is constant.
     
  5. Jun 26, 2015 #4

    Nathanael

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    By "vertical" the hint means "perpendicular to the ground" (not parallel to the velocity).
     
  6. Jun 26, 2015 #5
    so to solve i need to consider the 3rd spacial dimension?
     
  7. Jun 26, 2015 #6

    Simon Bridge

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    ... are there any forces in the third spacial dimension that could contribute to the "net force" that you are asked to find?

    "horizontal" and "vertical" are sometimes, lazily, used to refer to the orientation on the page - but not always.
    You have to use the context to determine what is intended.

    Since the officer moves in a circle - there is no particular reason to choose "vertically on the page" to be the direction of the velocity.
    You could have drawn the velocity diagonally on the page for eg. So the instructions cannot refer to the orientation on the page.

    The more precise meaning of "horizontal" is "parallel to the plane of the horizon" - and "vertical" means "perpendicular to the plane of the horizon" (the word means "aligned to the high/overhead vertex").
    You have correctly identified the important forces that are horizontal in this respect - in fact, the centripetal force is always horizontal so you can see how this is starting to make more sense?
     
  8. Jun 26, 2015 #7
    if the horizontal force in my diagram is the centripetal force then perpendicular to that is the velocity which would be in the vertical direction is my understanding.
     
  9. Jun 26, 2015 #8

    Nathanael

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    Whoever wrote the hint doesn't know how you drew your diagram, though. As Simon said, you could've easily (and correctly) drawn it at any angle.

    At any rate, the hint is talking about the other direction (perpendicular to both velocity and acceleration). Does the seat exert a force in that direction? If so, then that force must contribute to the net force.
     
  10. Jun 26, 2015 #9
    Yes the force the officer exerts on the seat is F=ma, (57.1kg)(-1.089m/s/s)=62.2N. But this is just the opposite of the centripetal force.
     
  11. Jun 26, 2015 #10

    Nathanael

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    Sorry, I misread and thought we were finding the force from the seat on the officer. But that is not the force I'm talking about. I said in the other (upwards) direction. Are there any forces in that direction? (The direction out of the page in your drawing.)
     
  12. Jun 26, 2015 #11

    Simon Bridge

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    No it isn't. Unless it's a very funny corner - think about the last time you went around a corner in a car.
     
  13. Jun 26, 2015 #12
    If i was in the car taking the turn the force of me on seat would be in the direction opposite the centripetal force, ignoring my weight on the seat.
     
  14. Jun 26, 2015 #13
    that would be the direction going out the right side of my diagram
     
  15. Jun 26, 2015 #14
    It would be a frictional force between me and the seat.
     
  16. Jun 26, 2015 #15

    Simon Bridge

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    ... what you are saying is that only one component of the net force of you on the seat has been calculated by you.
    What do you get if you do not ignore the weight?
     
  17. Jun 26, 2015 #16
    If the weight of me was added, then it would be a vector pointing in the -z direction?
     
  18. Jun 26, 2015 #17
    But i thought the question is asking for the force on the same axis as the centripetal force. Just in the opposite direction.
     
  19. Jun 26, 2015 #18

    Simon Bridge

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    What would be a vector pointing in the -z direction? Be specific!
    ... the question asks for the "net force" ... that is the vector sum of all the forces on the object.

    You correctly calculated the centripetal force - but this was the wrong answer ... therefore what you thought the question was asking must be wrong. Either that or the person who designed the assignment is wrong. Pick one. If you do not follow advise I cannot help you.
     
  20. Jun 26, 2015 #19
    sorry, my diagram is on the xy plane x axis being the axis parallel to the centripetal force, y axis is perpendicular to that. positive z axis perpendicular to the xy plane in the direction coming out of the page if your looking at it. and gravity would be a vector field perpendicular to the xy plane pointing in the negative direction.
     
  21. Jun 26, 2015 #20

    Nathanael

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    All the question asked for was the "net force." If they say "net force" you can be sure they want the sum of the forces in all directions.
     
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