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Circular Motion of a car Problem

  1. May 3, 2003 #1
    I'm kind of stumped on the following problem:

    A car rounds a banked curve. The radius of curvature of the road is R, the banking angle is theta, and the coefficient of friction is u.
    a) Determine the range of speeds the car can have without slipping up or down the banked surface.

    This is what I've done for when the car is on the verge of sliding up the embankment. According to the way I've done it, the equation for sliding down the embankment is the same.

    The sum of the forces in the radially inward direction is:
    (the sum of the normals on the right and left tires) (N1 + N2)sintheta + (force of friction)costheta = ma(radially inward) = mvtan/R.

    The sum of the forces in the vertical directions are:
    (N1 + N2)costheta - (force of friction)sintheta - mg = 0

    If I divide the first equation by the second I get:

    vtangent = (R(tantheta - cottheta))^1/2 so tantheta - cottheta >= 0, but when tantheta = cottheta, vtan = 0 and when cottheta = 0, tantheta goes to infinity, so my ranges for vtangent are zero to infinity.

    That can't be right. Can anyone tell me where I went wrong?

  2. jcsd
  3. May 4, 2003 #2
    I think you can do it this way:

    Total normal force is just N=m*g*cosT. Therefore the maximum force of static friction is u*m*g*cosT but in which direction?
    Centripetal force must always equal (m*v^2)/R. It is provided by a component of N directed toward the center of the circle: N*sinT=m*g*cosT*sinT.

    Now, what about the friction? At high speed, it acts along the slope downward, toward the center of the curve, so it should be added to the centripetal force provided by N. fcosT = u*m*g*(cosT)^2

    But at the lowest speed, friction acts UPWARD along the slope, counteracting the car's tendency to slide down, so that friction has a radial component acting outwards, away from the center of the curve. So, it should be subtracted from the net cetripetal force provided by N.
  4. May 4, 2003 #3
    Thanks for your reply.

    I see what you're saying, but if I divide the equation for the radially inward forces by the equations for the vertical forces I get the following:

    1)Radially inward forces car sliding down: NsinT - frcosT = mv^2/R
    2)Vertical forces car sliding down: NcosT + frsinT = mg

    Divide 1) by 2) and you get: v^2 = Rg(tanT - cotT)

    3)Radially inward forces car sliding up: NsinT + frcosT = mv^2/R
    4)Vertial forces car sliding down: NcosT - frsinT = mv^2/R

    Divide 3) by 4) and you get: v^2 = Rg(tanT - cotT)

    This result doesn't seem to have the "range of speeds" the problem's looking for. Instead, it seems to me that I've found the range of angles (pi/2 <= T < pi), but if you plug in these angles your velocity range is 0 to infinity.

    So anyway, I'm still not getting the desired results.
  5. May 4, 2003 #4
    I don't understand what you're trying to do by dividing those equations.

    I'm not 100% certain about this, but I would answer it this way:

    mg(cosT)(sinT) - umg(cosT)^2 <= m(v^2)/R <= mg(cosT)(sinT) + umg(cosT)^2

    the m's cancel out & you can multiply through by R to get:

    g(cosT)[sinT - u(cosT)] <= v^2 <= g(cosT)[sinT + u(cosT)]

    But as I said, I'm not sure, so I'd be happy to hear other opinions on this.
  6. May 4, 2003 #5
    Thanks Gnome.

    I don't know what I was thinking when I was doing that division. I saw another example where equations with single terms on each side were divided and for some reason, late at night, dividing these equations made sense for a little while.

    Anyway, NcosT - uNsinT = mg, so N = mg/(cosT - usinT) and if I plug this into the equation for the radially inward forces I get my two different equations for vtangential when the cars on the verge of sliding up and on the verge of sliding down.

    I believe this is correct. If it's not can someone please let me know.

    Thanks again for taking the time to look over my problem and post a reply.
  7. May 4, 2003 #6
    I think you're going off in the wrong direction again.
    Don't try to work the vertical forces or forces along the sloped surface. Just concentrate on balancing the centripetal (HORIZONTAL) forces, and resolve the normal and frictional forces into their horizontal components.
    Assuming we're both using
    N = total normal force
    f = maximum static friction between tires & road
    F = total centripetal force

    N = mg(cosT) (this acts perpendicular to slope)
    f = uN = umg(cosT) (this acts along the slope, upwards at low speed, downwards at high speed)
    f[h] = component of friction in HORIZONTAL direction = ??? (you fill this in)
    N[h] = component of normal force in HORIZONTAL direction = ??? (you fill this in)
    F = m(v^2)/R = sum of horizontal components of normal and frictional forces (ALWAYS!) but f is variable. You are looking for the two extremes: maximum f, positive and negative

    Put this all together & I think you will end up with the inequality I wrote earlier.
  8. May 5, 2003 #7
    I think I'm right this time about N. My equations for v agree with equations for v in similar problems that I've found.

    Thanks again.
  9. May 5, 2003 #8
    Got it.

    I was wrong to assume that N = mg(cosT).
    Actually, the equation is NcosT +/- uNsinT = mg (+/- depends on whether f is acting upward or downward).

    So, at low speeds (when f is acting UP the slope):
    NcosT + uNsinT = mg and
    mgcosTsinT - uNcosT = m(v^2)/R

    and at high speeds (when f is acting DOWN the slope):
    NcosT - uNsinT = mg
    mgcosTsinT + uNcosT = m(v^2)/R

    Solving those will give the maximum and minimum values of v.
  10. May 6, 2003 #9

    v^2= Rg(sinT - ucosT)/(cosT + usinT) and
    v^2= Rg(sinT + ucosT)/(cosT - usinT)
  11. May 6, 2003 #10
    Are you sure?
    I'm getting
    v^2 = Rg((cosT)^2)(sinT+ucosT)/(cosT-usinT)
    v^2 = Rg((cosT)^2)(sinT-ucosT)/(cosT+usinT)

    For example, the high-speed case is:
    NcosT - uNsinT = mg and
    mgcosTsinT + uNcosT = m(v^2)/R
    so N = mg/(cosT-usinT) and then substituting into the 2nd eqn...
    mv^2 = RmgcosTsinT + umgcosT/(cosT-usinT)
    v^2 = Rg[(cosTsinT(cosT-usinT) + ucosT)/(cosT - usinT)]
    = Rg[(((cosT)^2)sinT - ucosT(sinT)^2 + ucosT)/(cosT-usinT)]
    = Rg[(((1-sinT)^2)sinT + ucosT(1-(sinT)^2))/(cosT - usinT)]
    = Rg((cosT)^2)[(sinT + ucosT)/(cosT - usinT)]

    The steps for the other case are equivalent.

    How are you getting rid of the (cosT)^2 term?
  12. May 7, 2003 #11
    The sum of the centripetal forces is NsinT - uNcosT = mv^2/R not mgcosTsinT + uNcosT = m(v^2)/R. NsinT is the component of the Normal force in the centripetal direction.
  13. May 7, 2003 #12

    When I put the "N"s back I forgot that one.

    Thanks, d.
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