A child’s toy consists of a small wedge that has an acute angle θ. The sloping side of the wedge is frictionless, and an object of mass m on

Question

A child’s toy consists of a small wedge that has an acute angle θ. The sloping side of the wedge is frictionless, and an object of mass m on it remains at constant height if the wedge is spun at a certain constant speed. The wedge is spun by rotating, as an axis, a vertical rod that is firmly attached to the wedge at the bottom end. Show that, when the object sits at rest at a point at distance L up along the wedge, the speed of the object must be v = (gL sin(θ))1/2.

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Kiệt Gia 6 months 2021-07-31T13:11:23+00:00 2 Answers 44 views 0

Answers ( )

    0
    2021-07-31T13:13:05+00:00

    Answer:

    see explanation

    Explanation:

    Net force along x axis is

    \sum F_x = F \sin \theta= \frac{v^2}{R} ----(1)

    Net force along y axis is

    \sum F_y = F \cos \theta= mg ----(2)

    The object can along accelerate down the stamp.

    Thus F(net) is at the angle down the stamp

    F_{net}=F_c\\\\F_{net}= \sin \theta mg\\\\F_c = \frac{mv^2}{r} \\

    where r = L in the direction of acceleration

    \sin \theta mg = \frac{mv^2}{L}

    v^2 = gL \sin \theta

    v = \sqrt{gL \sin \theta}

    v = (gL \sin \theta )^{1/2}

    0
    2021-07-31T13:13:08+00:00

    Answer:

    It has been proved that;

    v = (gLsinθ)^(1/2)

    Explanation:

    Normal force is a force that provides support to an object when the object is in contact with another stable object. The normal force is perpendicular to the object and also known as the supporting force.

    Where;

    m is the mass of the object

    v is the speed of the object

    θ is the angle of wedge

    L is the distance between the position of the object and the bottom.

    The net force along the x-axis is;

    ΣF_x: = Nsinθ = mv²/r   – – – (eq1)

    Where N is the normal force.

    The net force along the y-axis is;

    ΣF_y: = Ncosθ = mg

    And N = mg/cosθ

    Let’s put mg/cosθ for N in eq(1)

    Thus,

    (mg/cosθ)•sinθ= mv²/r

    m will cancel out to give;

    (g/cosθ)•sinθ = v²/r

    v² = r(g/cosθ)•sinθ

    Now, from the free body diagram I’ve drawn, using trigonometric ratio, r = Lcosθ

    Thus,

    v² = (Lcosθ)(g/cosθ)•sinθ

    v² = gLsinθ

    Taking square root of both sides,

    v = √(gLsinθ)

    Or v = (gLsinθ)^(1/2)

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