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Centripetal force on a turning bus

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider the rear view of a bus turning right. If the driver turned the bus while travelling too fast, the bus could topple over. If the bus did topple would the bus fall to the left or to the right?

    Since the bus was turning to the right, the centripetal force is to the right. centripetal force =mv^2/r so if it is too fast, velocity would increase and centripetal force would increase and the bus would topple to the right. However, the answer to this question says:

    •The bus would tip to the left.
    • The force of friction between the tyres and the road provides the
    centripetal force at the base of the bus, providing an unbalanced
    moment about the bus’s centre of mass [1]
    • causing it to topple anticlockwise. [1]

    Can someone please explain this to me?
    Thank you!
    2. Relevant equations
    Fc = (mv^2)/r

    3. The attempt at a solution
    please see above
  2. jcsd
  3. Oct 4, 2009 #2
    Well the bus wants to keep going in a straight line, and if the acceleration that tries to force it into a new path isn't high enough, it will follow that straight path and topple over as the rest of it tries to keep turning.
  4. Oct 4, 2009 #3
    Isn't the bus going too fast so centripetal force that allows it to turn right is larger, hence providing a torque that will tip the bus over to the right (i.e. clockwise torque)?
  5. Oct 4, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    But where on the bus does the friction force (which provides the centripetal force) act? What torque does that force exert about the center of mass?
  6. Oct 4, 2009 #5
    That's the key to it all, the force is applied to one part of the bus, but not to the centre of mass.
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