To determine the coefficient of static friction between two materials, an engineer places a small sample of one material on a horizontal disk whose surface is made of the other material and then rotates the disk from rest with a constant angular acceleration of 0.4 rad/s2. If she determines that the small sample slips on the disk after 9.903 s, what is the coefficient of friction

This question is incomplete, the missing image is uploaded along this answer.

the coefficient of friction is 0.32

Explanation:

 Given the data in the question;

we make use of kinematic equation of motion;

ω = ω₀ + ∝t

we substitute

ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )

ω = 3.9612 rad/s

The centripetal force acting on the sample is;

Fc = mrω²

from the image; r = 200 mm = 0.2 m

so we substitute

Fc = m(0.2 m ) ( 3.9612 rad/s )²

Fc = (3.13822 m/s²)m

we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;

f = Fc

μN = Fc

μmg = (3.13822 m/s²)m

μ = (3.13822 m/s²)m / mg

μ = (3.13822 m/s²) / g

acceleration due to gravity g = 9.8 m/s²

so

μ = (3.13822 m/s²) / 9.8 m/s²

μ = 0.32

Therefore, the coefficient of friction is 0.32

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