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A coin is placed 35 cm from the center of a
horizontal turntable, initially at rest. The
turntable then begins to rotate. When the
speed of the coin is 110 cm/s (rotating at a
constant rate), the coin just begins to slip.
The acceleration of gravity is 980 cm/s^2
.
What is the coefficient of static friction between the coin and the turntable?

Respuesta :

MathPhys

Answer:

0.35

Explanation:

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Sum of the forces in the centripetal direction:

∑F = ma

Nμ = m v² / r

mgμ = m v² / r

gμ = v² / r

μ = v² / (gr)

μ = (110 cm/s)² / (980 cm/s² × 35 cm)

μ = 0.35

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