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