Ask Me Help Desk

Ask Me Help Desk (https://www.askmehelpdesk.com/forum.php)
-   Mechanical Engineering (https://www.askmehelpdesk.com/forumdisplay.php?f=494)
-   -   Tyre force (https://www.askmehelpdesk.com/showthread.php?t=796845)

  • Jul 16, 2014, 07:33 AM
    Mechsol61
    Tyre force
    If a car travelling in a 20 metre circle at 10 kilometres per hour increases its speed to 20 kilometres per hour does the force of the tyre on the road double or square ?
  • Jul 16, 2014, 08:09 AM
    ebaines
    Hint: the magnitude of centripetal acceleration for an object moving in a circle at constant speed is . Memorize it! Note the v (velocity) term, and recall that the force needed to keep the object moving in the circle is detemined from . So now what do you think the answer is?
  • Jul 16, 2014, 11:10 PM
    Mechsol61
    I haven't a clue. I'm an electrical engineer and that's why I'm asking the question.
  • Jul 17, 2014, 05:31 AM
    ebaines
    Think about the formulas I gave you. Centripetal acceleration goes as the square of velocity, and the force between the tires and the road to keep the car moving in a circle is proportional to the centripetal acceleration.
  • Jul 17, 2014, 05:48 AM
    Mechsol61
    Hi ebaines. So from that, I take it that when the speed doubles the force between the tyre and the road squares ! I know this happens to the wind resistance of a motor vehicle when its speed doubles the force of the air acting on it actually squares, but the laws of physiscs are something I don't use on a daily basis. Thank you for your assistance.
  • Jul 17, 2014, 05:52 AM
    ebaines
    Correct! Another pheneomenon that goes as velocity squared is kinetic energy: KE = (1/2)mv^2.

  • All times are GMT -7. The time now is 03:25 AM.