When a disk rolls on a surface without slip, the velocity of the point of contac

Question

When a disk rolls on a surface without slip, the velocity of the point of contact of the disk with the surface is zero and the friction force is less than the maximum allowable value of mu_k*N. The relative velocity equation is used to relate the velocity of the mass center to the angular velocity of the disk. a) What is the maximum value of P such that the disk rolls without slip? b) What is the angular acceleration of the disk for this value of P? The disk has a radius of 0.20m, and a mass of 1.8kg. Force P acts at the center of mass, and the coefficient of kinetic friction (mu_k) is 0.25. I know that the value of P is: 3/2 m*r*alpha. (alpha being the angular acceleration), and I know that the eventual expression for the max value of P is 3*mu_k*m*g. I also know the value of the ultimate acceleration from this particular P value = 24.5 rad/s^2. The problem I am having is how to show where the Pmax = 3*mu_k*m*g comes from. I need to see a derivation for this quantity.

Explanation / Answer

For disc to perform pure rolling not only velocity at point of contact should be zero but also acceleration at point of contact should be zero.that will be achieved when a = alpha*r now, a = (P- f)/m where f is friction alpha = ( f*r)/moment of inertia = (f*r)/(m*r^2/2) so, (P- f)/m = (f*r)/(m*r^2/2) * r so P = 3*f = 3*mu*mg

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