Physics Circular Motion...Please help/!!?

Question

A copper block rests 30 cm from center of a turntable. Coefficient of static friction between surface and block is 0.53. Turntable starts from rest and rotates at a constant angular acceleration of 0.5 rad/s^2. At what time wil block start to slip? (hint: normal force= weight of block)

Answer 1

You want to find the time when the force that friction can provide to keep the block in the circular path is equal to, and about to become bigger than, the required centripetal force.

Set up the equations as follows:

u = coefficient of static friction

u*m*g=m*a

cancel the masses, and substitute (V^2)/R for a

u*g=(V^2)/R

Now, recall that Vfinal=At

u*g=(a^2)*(t^2)/R

Now, solve for t and plug in the variables.

sqrt[u*g*R/(a^2)]=(t^2)

Answer 2

As the table turns, the only force that prevents the block from slipping radially (outward) is static friction. The max static friction is u*mg (u= 0.53). The radial force (centripetal or centrifugal is up to your conceptual understanding) is m* r * w (w is angular velocity). Distance r is 30 cm but should be 0.3 meter to be consistent with SI unit (because g is m/s*s)

w = w0 + a t (where a is angular acceleration). w0=0

Force equation

m * r* (a*t) * (a*t) = u*m*g

Mass m cancels on both sides, so

t*t = u /(r * a* a), you do the square root to get the answer.