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DESPERATE please help me! Circular motion

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution

    I only know how to do this with the coefficient of friction, and since there is no friction here I don't know how to do it. Normally the equation is =
    Fstatic friction max = v^2/r
    mu * g * r = v^2
    and then finally
    root(mu*g*r) = v

    What do I do if I don't have mu?
     
    Last edited: Dec 16, 2008
  2. jcsd
  3. Dec 15, 2008 #2

    LowlyPion

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    Draw a force diagram.

    You have gravity acting vertically.

    You have centrifugal force acting outwardly and horizontally.

    The downward force of gravity along the banked incline must be balanced by the upward along the bank component of the centrifugal force shouldn't it?
     
  4. Dec 15, 2008 #3
    I have no clue how to find centrifugal force, if you would provide me with the sine/cosine explanation I'd be forever grateful.

    I did draw a force diagram. Is the pull of gravity sine(14) then you multiply this by something to get the Force?
    I would be extremely grateful if you'd post a little guide to the problem, don't even use my numbers, I just really need to get this do quickly. Unfortunately I have a lot of other problems to go through. I'm really desperate I have to get this finished and I'm not sure how to do it. If I get the answer tonight I can ask my teacher tomorrow, I just really need the credit for the check off tomorrow.
    How do I figure out centripetal force in this case? Would you at least give me the equation for this in this situation? i know I need to use sine and cosine but I'm not sure how.
     
  5. Dec 15, 2008 #4

    LowlyPion

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    m*g*sin14 is the downward force along the bank.

    (m*v2*cos14)/r is the upward force.
     
  6. Dec 15, 2008 #5
    Thank you so much. So the m*g*sin14 is my centripetal force?
     
  7. Dec 15, 2008 #6
    "The downward force of gravity along the banked incline must be balanced by the upward along the bank component of the centrifugal force shouldn't it?"

    I agree with this, but all this means is that the vertical component of R is 8500N. I just don't see how this ties in to finding the maximum speed of the car. I'm confused between how the problems link together.
     
  8. Dec 16, 2008 #7

    rl.bhat

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    To keep a car in a circular orbit with uniform velocity, a centripetal force must act on the car. If R is the normal reaction force on the car, its herizontal component i.e. R*sin(theta) will provide the necessary centripetal force. The car is not moving in the upward direction, because the vertical component of R, i.e. R*cos(theta) is balanced by the weight of the car, i.e. mg. Since R is not given in the problem, we can eliminate from the problem by wrighting
    R*sin(theta)/R*cos(theta) = (mv^2/r)/mg
    or tan(theta) = v^2/rg
    Now you can the velocity.
     
  9. Dec 16, 2008 #8

    LowlyPion

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    The 8500N looks like the weight (m*g).

    If you set the 2 terms equal to each other then all you don't know is v and happily that's all they are asking.

    m*g*sin14 is gravity

    (m*v2Cos14)/r is the centrifugal force (acting outward against the bank)
     
  10. Dec 16, 2008 #9
    Ok, I actually think I understand it now. I was just confusing myself with the angles. This is so similar to the problems we did in dynamics I feel kind of stupid now.

    We were talking about before how the horizontal component of the force has to be sin(14) * 8500. I understand that bit but I was reading on and I have a question (on for this same setup) asking me to:
    "stat the magnitude and the direction for the resultant force acting on the car"
    Is this just asking for R? I'm not sure what the "resultant force" is
    And if it is R would I say its perpendicular to the slanted road?
     
  11. Dec 16, 2008 #10
    "The 8500N looks like the weight (m*g).

    If you set the 2 terms equal to each other then all you don't know is v and happily that's all they are asking.

    m*g*sin14 is gravity

    (m*v2Cos14)/r is the centrifugal force (acting outward against the bank)"

    But Rl.Bhat's explanation works as well right? I'm a little confused about what your saying (mainly since my teacher said that centrifugal force is a percieved force and what I'm actually looking for is centripetal force)
     
  12. Dec 16, 2008 #11

    LowlyPion

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    The reaction force R is merely the components of the forces that are normal to the bank.

    m*g*cos14 and mv2/r*sin14 added together.
     
  13. Dec 16, 2008 #12

    LowlyPion

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    Yes centripetal acceleration is acting on the car. Centrifugal is the perception of that force. I was just trying to make it easier for you to see.

    rl.bhat's results in exactly what those reduce to. Sin14/Cos14 is Tan14
     
  14. Dec 16, 2008 #13
    Oh alright then, that makes sense to me.
    Do you understand what the other question is asking about "state the magnitude and direction of the resultant force acting on the car"

    What is this resultant force?
     
  15. Dec 16, 2008 #14

    rl.bhat

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    Since there is no friction, the net force acting on the car is centripetal force.
    Find the mass of the car. And substitute in m*v^2/r.
     
  16. Dec 16, 2008 #15
    to find mass 8500N / 9.81 = 866 kgs
    so I multiply that by V^2/r
    The problem is that when this question is asked, they have not yet provided the radius so I don't have that, nor do I have the velocity
     
  17. Dec 16, 2008 #16

    rl.bhat

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    O.K.
    Now R*cos(theta) = mg. Find R.
    Net force is R*sin(theta)
     
  18. Dec 16, 2008 #17
    Right, I understand that bit but if Fnet=R*sine(14) then I need to use the R value from above right? So using the two sides I know 8500N and then 2100N (the rounded up horizontal component of R) so I use the Pythagorean theorum right? a^2 + b^2 =C^2 to find the hypotenuse?
     
  19. Dec 16, 2008 #18

    rl.bhat

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    8500N and R*cos(14) are equal and opposite. So the cancel each other. Only unbalanced force is R*sin(14). And that is F(net). So you need not use Pythagorean theorem.
     
  20. Dec 16, 2008 #19

    rl.bhat

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    Well. If the car is at rest, the forces acting on it will be 8500N and 8500*cos(14) = 8248 N and angle between them is (180 - 14 ) degree. If you know the vector addition you can find the magnitude and direction of the resultant force on the car.
     
  21. Dec 16, 2008 #20
    Alright thanks a lot. I really appreciate the time you spent helping me.
     
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