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A small cylinder rests on a circular turntable that is rotating clockwise at a constant speed. The cylinder is at a distance of r = 12 cm from the center of the turntable. The coefficient of static friction between the bottom of the cylinder and the surface of the turntable is 0.45. What is the maximum speed vmax that the cylinder can have without slipping off the turntable?

Respuesta :

Answer:

The maximum speed vmax = 0.727 m/s

Explanation

Given that

Distance ,r= 12 cm

Coefficient of static friction , μ= 0.45

Lets the mass of cylinder is m.

The friction force on the cylinder Fr = μ m g

Force due to rotation F

[tex]F=mv^2/r[/tex]

v= Speed of rotation

The condition for no slipping

Fr= F

[tex]\mu mg=mv^2/r[/tex]

[tex]v=\sqrt{\mu rg}[/tex]

Now by pitting the values

[tex]v=\sqrt{0.45\times 0.12\times 9.81}\ m/s[/tex]

v=0.727 m/s

So

The maximum speed vmax = 0.727 m/s

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