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weqwewe [10]
3 years ago
13

A 4.20 gg coin is placed 15.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with

the turntable surface of μsμsmu_s = 0.700 and μkμkmu_k = 0.420. Part A What is the maximum angular velocity with which the turntable can spin without the coin sliding?
Physics
1 answer:
ololo11 [35]3 years ago
5 0

Answer:

Explanation:

mass of the coin m = 4.2 x 10⁻³ kg

radius of circular path r in which coin will move = 15 x 10⁻² m

coefficient of static friction μs = .7

at maximum angular velocity of ω , limiting force of friction will provide centripetal force to coin to rotate in circular path .

so centripetal force = frictional force

mω²r = μs x mg

ω²r = μs x g

ω² = μs x g / r

= .7 x 9.8 /  15 x 10⁻²

= 45.73

ω = 6.76 radian / s.

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