Physics

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A coin is placed 17.0 from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 33 is reached and the coin slides off.What is the coefficient of static friction between the coin and the turntable?

  • Physics -

    forcefriction=forcecentripetal
    mu*mg=m*v^2/r

    At this point, I don't know what your 33 is? m/s? rpm?

  • Physics -

    oops sorry, 33 rpm and 17 cm

  • Physics -

    I have a very similar problem due on mastering physics in an hour.. I came here to get help but now 30 minutes later I'm actually going to be giving help :). Hopefully it wont be too late depending on your time zone.
    u= coefficient of static friction
    force of friction = u*mg

    IM NOT THAT SMART SO THIS MIGHT BE WRONG BUT HOPEFULLY IT WILL HELP.. I ALSO SUCK AT SIG FIGS SO ITS PROBABLY SLIGHTLY OFF.

    set force of friction = to centripetal force

    so u*m*g = m*v^2/r

    m cancels

    so u*g = v^2/r.
    r = 17cm = .17m

    the circumference of the circle = 2pi*r
    which roughly = 1.068m (only 2 sig figs tho)

    convert 33 rpm to rps to get .55 rps.

    so the object move .55 rotations in one second so it moves .55*1.068m in one second. so it moves .5874m/s reduce this to 2 sig figs to get .59m/s

    use the equation we formed earlier :
    u*g = v^2/r.

    u= v^2/rg
    u= (.59m/s)^2/((.17m)(9.80m/s^2))
    u=.21

  • Physics -

    THANK YOU!!!!!!!

  • Physics -

    TY! TY!

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