Classical Mechanics - Circular Motion

The coefficient of static friction between a small coin and the surface of a turntable is 0.30. The turntable rotates at 33.3 revolutions per minute. What is the maximum distance from the center of the turntable at which the coin will not slide?

A. 0.024 m
B. 0.048 m
C. 0.121 m
D. 0.242 m
E. 0.484 m

(GR0177 #02)

Solution:

The coin will not slide if  F_centripetal = F_friction
mω²r = μ_smg
r = μ_sg/ω² =  μ_sg/(4π²f²)

Given:
μ_s = 0.30
f = 33.3 revolutions per minute = 33.3/60 = (100/3)(1/60) = 5/9
and take π² = (3.14)² ≈ 10

Radius:
r = (0.3)(10)/[4(10)(25/81)] = (3)(81)/[4(10)(25)] = 243/1000 = 0.243

Answer: D


Notes:


F_centripetal = mv²/r = mω²r
F_friction = μ_sN = μ_smg