Reham asked in Science & MathematicsPhysics · 7 months ago

A coin lies on the top of a turntable at a distance R from the center?

A coin lies on the top of a turntable at a distance R from the center. The table rotates at a

constant speed v. What is the minimum coefficient of static friction that can prevent the

coin from sliding of the table?

3 Answers

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  • oubaas
    Lv 7
    7 months ago

    m*g*μ ≥ m*V^2/r

    mass m cross

    μ ≥ V^2/(r*g)

  • 7 months ago

    When the turntable is rotating, there is centripetal force on the coin,

    Fc = m * v^2 ÷ R

    To prevent the coin from sliding off of the turntable, the friction force must be equal to the centripetal force,

    Ff = μ * m * g

    μ * m * g = m * v^2 ÷ R

    μ = v^2 ÷ (g * R

    I hope this is helpful for you.

  • 7 months ago

    Centripetal force F = mV²/R

    which is force of friction

    weight of the coin is mg

    coefficient of static friction = friction force/weight

    µ = mg / mV²/R = gR/V²

    edit, correction

    µ = (mV²/R) / mg = V²/gR

    Centripetal force f = mV²/r = mrω²

    ω is angular velocity in radians/sec

    1 radian/sec = 9.55 rev/min

    m is mass in kg

    r is radius of circle in meters

    V is the tangental velocity in m/s = ωr

    f is in Newtons

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