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A block is resting on a platform that is rotating at an angular speed of 2.5 rad/s. The coefficient of static friction between the block and the platform is 0.78. Determine the smallest distance from the axis at which the block can remain in place wothout skidding as the platform rotates.

Respuesta :

Answer:

[tex]r = 1.22 m[/tex]

Explanation:

As we know that block will not slip on the rotating disc then we will have

Friction force = centripetal Force

so here we will have

[tex]F_f = m\omega^2 r[/tex]

Here we know that

[tex]F_f = \mu N[/tex]

so we have

[tex]N = mg[/tex]

[tex]F_f = \mu mg[/tex]

[tex]m\omega^2 r = \mu mg[/tex]

[tex]r = \frac{\mu g}{\omega^2}[/tex]

[tex]r = \frac{0.78\times 9.81}{2.5^2}[/tex]

[tex]r = 1.22 m[/tex]

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