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A coin of mass m is on a rigid disk at a distance d from the center of the disk. There is friction between the coin and the disk. The coefficient of static friction is μs. At time t = 0, the disk begins to rotate with a constant angular acceleration of magnitude α. The magnitude of the acceleration due to gravity is g. Express you answers in terms of some or all of the given variables m, d, μs, α, t and g as needed. (a) While the coin remains at rest relative to the disk, what is fs, the magnitude of the force of static friction exerted by the disk on the coin as a function of time t? (b) At what angular speed ω will the coin start to slip with respect to the disk?

Respuesta :

lcmendozaf

Answer:

a) [tex]fs = m*d*\alpha^{2}*t^{2}[/tex]

b) [tex]\omega =\sqrt{\frac{\mu s*g}{d} }[/tex]

Explanation:

If the does not slip frm the surface, the friction force will equal to the centripetal force:

[tex]fs = m*Fc =m*\frac{V^{2}}{d} =m*\frac{\omega^{2}*d^{2}}{d}=m*d* \omega^{2}[/tex]    Let's call this, eq1.

We also know that angular acceleration is constant, so:

[tex]\omega = \alpha*t[/tex]   Replacing this on the fs formula:

[tex]fs = m*d* \alpha^{2}*t^{2}[/tex]

For the maximum speed before the coin slips, the friction will be [tex]fs = \mu s * N[/tex]  and we know that N=m*g. Replacing this values into our previous eq1 result leads to:

[tex]\mu s *m*g = m*d* \omega^{2}[/tex]

Solving for \omega:

[tex]\omega =\sqrt{\frac{\mu s*g}{d} }[/tex]