I really need help PLZ? ( question regarding revolutions)?

Question

A small turtle placed on a horizontal, rotating turntable at distance of 16.2 cm from its center. dizzy mass is 66.9g and the coefficient of static friction between his feet and the turntable is .445 ......the acceleration of gravity is 9.8 m/s2

a) find the maximum number of revolutions per second the turntable can have if the turtle is remain stationary relative to the turntable answer in rev /s

b) the turn table starts from rest at t=0 and has a uniform acceleration of 4.62 @ t=0 and has a uniform acceleration of 4.62 rad/s2.. find the time at which turtle begins to slip answer in sec

Answer 1

r = 0.162 m

m = 0.0669 kg

g = 9.8 m/s^2

μ = 0.445

amax = (0.445)(9.8 m/s^2) = 4.361 m/s^2

a)

with no angular acceleration

amax = rω^2, ω in rad/sec

ω ≈ √[(4.361 m/s^2)/(0.162 m)]

ω ≈ (5.1884 rad/sec)(1 rev/2π rad.)

ω ≈ 0.82576 rev/s ≈ 0.826 rev/s

b)

a^2 = (rα)^2 + (rω^2)^2

ω = 0 + αt

a^2 = (rα)^2 + (r(αt)^2)^2

(r(αt)^2)^2 = a^2 - (rα)^2

r(αt)^2 = √[a^2 - (rα)^2]

t = √{(1/rα^2)√[a^2 - (rα)^2]}

t ≈ √{(1/((0.162 m)(4.62 rad/s2)^2))√[(4.361 m/s^2)^2 - ((0.162 m)(4.62 rad/s2))^2]}

t ≈ √{(1 s^4/(3.4577928 m))√[19.018321 m^2/s^4 - 11.95633104773184 m^2/s^4]}

t ≈ √{(1 s^4/(3.4577928 m))√[7.06198995226816 m^2/s^4]}

t ≈ √{(2.65744049 s^2)/3.4577928}

t ≈ √(0.7685366 s^2)

t ≈ 0.8766622 s ≈ 0.877 s