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    Mechsol61's Avatar
    Mechsol61 Posts: 3, Reputation: 1
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    #1

    Jul 16, 2014, 07:33 AM
    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 ?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Jul 16, 2014, 08:09 AM
    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?
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    Mechsol61 Posts: 3, Reputation: 1
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    #3

    Jul 16, 2014, 11:10 PM
    I haven't a clue. I'm an electrical engineer and that's why I'm asking the question.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Jul 17, 2014, 05:31 AM
    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.
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    Mechsol61 Posts: 3, Reputation: 1
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    #5

    Jul 17, 2014, 05:48 AM
    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.
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    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Jul 17, 2014, 05:52 AM
    Correct! Another pheneomenon that goes as velocity squared is kinetic energy: KE = (1/2)mv^2.
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