Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 7.6 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip? (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.16 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.

Answer 1

a)

w = angular speed of turntable = 33 rev/min = 33 (2pi)/60 = 3.5 rad/s

r = distance of seed from the axis of rotation = 7.6 cm = 0.076 m

acceleration of the seed is given as

a = r w² = (0.076) (3.5)² = 0.931 m/s²

b)

Coefficient of static friction is given as

u_s = a/g = 0.931/9.8 = 0.095

c)

tangential acceleration is given as

a_t = r alpha = r (w - 0)/t = rw/t = 0.076 x 3.5/0.16 = 1.6625

centripetal acceleration is given as

a_c = rw² = (0.076) (3.5)² = 0.931 m/s²

Net acceleration is given as

a = sqrt(a_c² + a_t²) = sqrt(( 0.931)² + (1.6625)²) = 1.91

Coefficient of static friction is given as

u_s = a/g = 1.91/9.8 = 0.195