Physics question about Rotational Motion and the Law of Gravity:?

Question

A copper block rests 30.0cm from the center of steel turntable. The coefficent of static friciton between the two is 0.53. The turntable start from rest and rotates with a constant angular acceleration of 0.50rad/s^2. After what time internal will the block start to slip on the turntable? (Hint: The normal force in this case equals the weight of the block.)

Answer 1

The total acceleration to cause the block to slip is .53g, or 5.194 m/s^2.

The tangential acceleration is alpha*r = .50*.30 = .15 m/s^2 = At
The radial acceleration is r*w^2, where w = .5(alpha)t^2, so Ar = .30*.5*.50*t^2 Ergo, we have

5.194 = root(.15^2 + .005625t^4) Solving for t,

t = 5.506 sec

Answer 2

The available friction is 0.53*m*g = 5.2*m meters/s^2. That friction must provide the centripetal force to keep the block moving in a circle, m*v^2/r.

So set the 2 forces equal to find the limiting speed.
5.2*m meters/s^2 = m*v^2/r The term m for mass cancels out. Now solve for v^2
v^2 = r*(5.2 meters/s^2)
Because I used g = 9.8 m/s^2 above, I need to convert the radius to meters - so
v^2 = .3*5.2 meters^2/s^2 and find v
v = 1.2 m/s
That's linear speed, need to convert to radians per second

1.2 m/s*(2*pi*radians/2*pi*r) = 4.2 radians/s. At an acceleration of.50rad/s^2, it will reach that speed in t seconds per the equation
w = alpha*t