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?

Question

Jesse Lutz

Physics

29 April, 01:25

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?

+3

Answers (1)

Know the Answer?

Jaqueline Stout · Answer 1

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