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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.

Respuesta :

Olajidey

Answer:

see explanation

Explanation:

Net force along x axis is

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

Net force along y axis is

[tex]\sum F_y = F \cos \theta= mg ----(2)[/tex]

The object can along accelerate down the stamp.

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

[tex]F_{net}=F_c\\\\F_{net}= \sin \theta mg\\\\F_c = \frac{mv^2}{r} \\[/tex]

where r = L in the direction of acceleration

[tex]\sin \theta mg = \frac{mv^2}{L}[/tex]

[tex]v^2 = gL \sin \theta[/tex]

[tex]v = \sqrt{gL \sin \theta}[/tex]

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

onyebuchinnaji

The relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].

The given parameters:

  • inclination of the wedge, = θ
  • let the constant speed = v

The relationship between the distance of the object, speed of the object can be determined by the net force on the toy is calculated follows;

[tex]Fsin(\theta) = \frac{mv^2}{L} \\\\mgsin(\theta) = \frac{mv^2}{L} \\\\gsin(\theta) = \frac{v^2}{L} \\\\v^2 = gL sin(\theta)\\\\v = \sqrt{gL sin(\theta)}[/tex]

Thus, the relationship between the distance of the object and speed of the object is [tex]v = \sqrt{gL sin(\theta)}[/tex].

Learn more about speed of object in vertical circle here https://brainly.com/question/19904705

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